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THE CALCULUS 






THE CALCULUS 



(REVISED) 



An Elementary Treatise on the Differen 

tial and Integral Calculus, with 

Practical Applications 



Prepared for the Use of 

The Midshipmen of the United States 

Naval Academy 



STIMSON J. BROWN 

Professor of Mathematics , United States Naz-y 



PAUL CAPRON, A. M. 

Instructor, United S'ates Naval Academy 



1912 



QA 30 3 



Copyright, 1912, by 
STIMSON J. BROWN and PAUL CAPRON 



BALTIMORE, MD. , U. S. A. 



©C!.A30i>788 



CONTENTS. 

CHAPTER I. 

AT>m Derivatives and Differentials. „.„■ 

ART. PAGE 

1-5. General problem of the tangent 1 

6. Variation of functions 5 

7-10. Variable velocity — speed 7 

11-14. Derivatives — differentials 11 

15-16. Rules for differentiating — algebraic functions 14 

17-18. Derivative of implicit function 18 

19-23. Further differentials 19 

24-27. Differentials of trigonometric functions 24 

28-30. Differentials of inverse trigonometric functions 29 

31-36. Differentials of exponential and logarithmic functions. 32 

37. Problems in speed and time-rates 37 

CHAPTER II. 

Analytic Geometry. 

38-49. Recapitulation of graphic analysis 45 

50. Transformation of coordinates — shifting the origin and 

rotation of axes 49 

53-54. General equation of a conic 51 

55-56. The parabola 53 

57-63. The central conies 55 

64. Conjugate diameters 60 

65-66. Focal radii of central conies 63 

67-70. Geometrical applications of derivatives 64 

71-72. Properties of tangents to conies 72 

73. Pedal Curves 74 

74. Examples 76 

75-77. The second derivative — inflections 75 

78-80. Curvature — radius of curvature 82 

81-82. Differential of arc 84 

83-85. Curvature in rectangular coordinates 86 

86. Auxiliary circles of the ellipse 89 



vi Contents. 

ART. PAGE 

87-88. Parametric equations and their derivatives 91 

89-91. The cycloid 94 

92-94. Polar coordinates 100 

95-98. Derivatives and their applications in polar coordinates. 103 

99. Curve tracing — higher degree curves 106 

100-103. Approximate forms at origin and at infinity 107 

104-107. The analytical triangle 114 

108-111. Examples 117 

CHAPTER III. 

Maxima and Minima. 

112-113. Maxima and minima 122 

CHAPTER IV. 

Integration. 

114. Definition and table of elementary direct integrals 129 

115-117. Direct integration 131 

118-120. Integration by substitution — rational fractions 134 

121-123. Areas by integration 136 

124-125. Definite integrals 140 

126-128. Areas— sign of definite integral 144 

129-130. Volumes of revolution 147 

131-132. Integration by parts 149 

133-138. Integration of trigonometric functions 150 

139-142. Formulae of reduction 154 

143-149. Indefinite integrals of powers of trigonometric functions. 159 

150-153. Definite integral as the limit of a sum 164 

154-158. Element of integration and application to problems.... 170 

CHAPTER VI. 

Space Coordinates. 

159. Space Coordinates 177 

160-162. Rectangular and cylindrical coordinates 177 

163. Spherical coordinates 179 

164-165. Equations in three dimensions 179 

166. Analysis of equations by plane sections 181 



Contexts . vn 

CHAPTER VII. 
4M , Areas, Volumes, Arcs, axd Surfaces. _.„_ 

ART. fA-i^t, 

170-171. Collection of formulae for areas 185 

172-173. Sectorial areas by y = mx method 187 

174-175. Areas in polar coordinates 189 

176-177. Volumes of revolution 191 

178-179. Successive integration 193 

180-181. Volumes of revolution — polar coordinates 196 

182-184. Volumes by parallel sections 198 

185-186. Cylindrical volumes 203 

187. Figures of elements of area and volume 205 

188-191. Length of arc — surfaces of revolution — cylindrical sur- 
faces 205 

192-193. Other curved surfaces 210 

194. The loxodrome 121 

195-198. Mercator's projection of the terrestrial sphere and 

spheroid 213 



CHAPTER VIII. 

Series. 

199. Series 218 

201-204. Development of series — Maclaurin's series 218 

205-206. Approximate integration 221 

207. Differential equations 222 

208-209. Elementary series — logarithmic series 224 

210-212. Taylor's series — finite differences 225 

213-214. Small changes in the astronomical triangle 229 

215-217. Simpson's first rule 232 

218-222. Simpson's rules 235 

223-228. Evaluation of illusory forms 239 

CHAPTER IX. 
Meax Values. 

229-230. Mean values 245 

231. Illustrative examples 247 

232-233. Methods and examples 248 



viii Contents. 



CHAPTER X. 

art. Kinematics. pAGE 

234-238. Composition and resolution of displacements 251 

239. Examples 256 

240. Speed 256 

241-244. Velocity 257 

245. Examples 261 

246-251. Acceleration 262 

252. Examples 268 



CHAPTER XI. 

Forces. 

253-254. Nature of forces 270 

255. Law of motion 271 

256. Units of mass, force and acceleration 271 

257-259. Composition and resolution of forces 274 

260. Equation of motion for a given direction 276 

261. Examples 277 



CHAPTER XII. 

Motion of a Heavy Particle. 

262. Definitions 279 

263-264. Rectilinear motion under gravity alone 280 

265. Parabolic motion under gravity alone 282 

266. Rectlinear motion under any constant force 284 

267. Examples 285 

268-280. Rectilinear motion under two or more forces and under 

variable forces 286 

269. Hooke's law 287 

270. The force of gravity 288 

271. Examples 289 

272-274. Resistance of a rough surface — friction 291 

275-277. Atmospheric resistance 295 

278-279. Mechanical connections 297 

280. Examples 299 



Contexts. ix 

AET. PAGE 

281-286. Simple pendulum and variation of gravity 301 

287. Motion in a vertical circle 306 

288-289. Motion in a horizontal circle 308 

290. Examples 309 

CHAPTER XIII. 

Momentum, Work axd Energy. 

291-292. Mean speed under constant force 311 

293-297. Momentum, impulse and impact 311 

298-303. Work 316 

304-307. Work and energy 319 

308-310. Power 323 

311. Characteristics of motion 326 

CHAPTER XIV. 
Rigid Bodies. 

312. Mass of a body of variable density 327 

313. Resultant of like parallel forces 328 

314-315. Moments 329 

316-320. Parallel forces 331 

321-322. Centers of gravity 333 

323. Work done by gravity on an extended body 335 

324-325. Computation of centers of gravity — examples 336 

326-327. Theorem of Pappus — examples 339 

328. Wind pressure on a plane surface 342 

329-334. Fluid pressure — center of pressure 342 

335. Kinetic energy of an extended body 347 

336-339. Moment of inertia and radius of gyration 348 

340-346. I and k 2 with reference to perpendicular and parallel 

axes 351 

347-348. 7c 2 for a solid of revolution — examples 358 

349-350. k 2 for solids about any axis 360 

351-352. D'Alembert's principle 361 

353. The equation of rotation 364 

354-355. The compound pendulum 365 

356. k 2 determined by experiments 367 

357-358. Combined translation and rotation 368 



THE CALCULUS. 



CHAPTER I. 

The Differential Calculus. 

1. The treatment of functions in the fourth-class algebra 
showed how to find the slope of a graph by determining the de- 
rivative of the corresponding function, or to find, by the same 
means, the tangents to certain curves when the equations of the 
curves were given. These problems exemplify in an elementary 
manner the important properties of the Differential Calculus; 
we now have to extend the same principles further, and must 
adopt a more generally useful style of notation. 

In order to introduce this notation as simply as possible, we 
shall repeat part of last year's work with some variations. 

2. Increments. — In finding the slope of the graph of the func- 
tion y — x 2 , we supposed x to be increased to the value (x + h), 
y thereby being increased to the value (y + Jc). Either h or k 
might be negative, indicating a decrease. Such a change in a 
variable is called an increment, and two increments like these, 
one of which results as a consequence of the other, are called 
simultaneous increments. 

We shall now indicate any increment by a composite symbol 
consisting of the letter A (the Greek capital delta, equivalent to 
the English D) followed by the symbol for the variable of which 
it is an increment. Thus the increment h given to x to increase 
its value to (x+h), will now be written Ax and read " delta x," 
or " increment given to x." Similarly for Tc as used here we shall 
write Ay (delta y). 
2 



2 The Calculus. 

An increment such as Ax or Ay may be positive or negative, 
indicating an increase or a decrease. 

To represent a certain sort of increment, the use of which will 
be explained in the next article, the small Eoman d is used in 
place of the capital A. The resulting symbols dx, dy, are read 
d-x, d-y, or differential of x, differential of y. 

The following problem illustrates the use of these symbols for 
increments. 

3. Problem of the Tangent. —To draw a tangent at the point 

x 2 
P (x ,y ) of the parabola, y= — , referred to the pair of per- 

a 

pendicular axes XOY. (Fig. 1.) 




Fig. 1. 



As the point P of the tangent is given, it will be sufficient to 
find the slope of the tangent. 

Suppose the tangent to be the line P Q . Through P draw 



The Differential Calculus. 3 

P N parallel to OX of any convenient length dx, and through N 
draw a parallel to OY, meeting the curve at A and the tangent 
at Q . Call NQ = dy; then our problem is to find the value of 
dy, so as to get the ratio, dy/dx, of dy to the assumed length dx. 
This ratio will be the required slope of the tangent. 

Suppose P to be any point of the curve ; draw PM parallel to 
OY, meeting PJSf at M, and the secant P P, meeting AN at Q. 

Then P M=Ax and MP— Ay are increments which, if given 
to the coordinates (x 0f y ) of P , will change them to the co- 
ordinates (x + Ax, y + Ay) of P. If the point P is supposed to 
slide along the curve through P , the secant P P will occupy the 
position of the desired tangent when P passes through the point 
P ; during this process the point Q will move along the line AN, 
passing through Q when P passes through P . The desired 

slope of the tangent, p° = -^~ , is thus the value taken on by 

the variable ratio p N when P reaches P and Q reaches Q . 

ON PM Av 

But by similar triangles, jz N = " = -^ , and when P 

reaches P , Ax=0. Then the required slope of the tangent is 

A// 



dy 
dx 



Ax 



AX=0 

But now, since P is on the curve, 



and since P is on the curve, 
subtracting, we find 






to 



2x Ax+(Axy 
u a 9 





The Cal 


CULUS. 


Ay __ 2x + Ax . 

Az ~ a ' 


dy 


2x + Ax 




dx 


a 


Acc=0 


dv=-?-^- • ^/r 




\*y — 


a 



2x 



Thus, in order to draw the tangent, after laying off P N any 

2x 
convenient distance dx to the right, we lay off NQ , — - times as 

far upward, and draw P Q Q , the required tangent. 

If the tangent P Q makes the angle r with the axis OX, then 
evidently 

-^ = tanr = slope of tangent to curve. 

As P may be any point of the curve, the result is expressed 
in general : The tangent line at any point (x, y) of the parabola 

y — — has the slope — - . 
u ft r a 

x 2 
For instance, the slope of the tangent to y = — at the point 

2v3 

(3, f ) is — ~ — = 3 ; so that, if any distance dx is measured to 

the right of the point (3, f ) and three times as great a distance 
upward from the point thus reached, a point of the tangent will 
be found. 

Note the convenience in the discussion above of having two 
expressions for an increment; the A forms have been used for 
variable increments, and the differential, or d forms, for the 
fixed increments the ratio of which is to be determined. More- 
over, Ay is an increment of the ordinate of the curve, whereas dy 
is an increment of the ordinate of the tangent. 



The Differential Calculus. 



4. Equation of the Tangent. — For any curve except a straight 

line, -^- — tan r, the slope of the tangent, has different values for 

different points of the curve; its value for any particular point 
(x ,y ) is indicated by 



dy 
dx 



:=tan<r n . 



so.vo 



Since the equation of the line through (x , y ) having the slope 
m is y — y = m(x — x ) (Algebra, Art. 101), the equation of the 
tangent to a curve at the point (x , y ) of the curve is 



y-yo={v-Xo) 



dy 



dx 



*<h Vo 



5. Examples. 

1. Find the general expression for ~^- 



Ay 



Ax 



for the 



A£=0 



curve Sy = x 3 , and using 1 inch as unit, plot the points of the 
curve for which x— — 3, — 2, —1, 0, 1, 2, 3 respectively, and at 
each point construct the tangent, using any convenient value for 
dx. Sketch the curve. 



2. Derive the general expression for -^- = 



Ay 



Ax 



for each 



Aa?=0 



of the following curves, and write the equation of the tangent to 
each at the point indicated : 

(a) x 2 = Sy at (4, 2) ; (b) x 2 = 27y at (9, 3) ; (c) ±x 3 = y at 
(i, i) ; (d) Sy = a* at (2, 1) ; (e) ay 2 = b 2 x at (x , y ). 

Ans. (a) x-y = 2, (b) 2x-3y = 9,- (c) 3x-y=l, (d) 
3s-2y=4, (e) 2ay y=l) 2 (x+x ). 

6. Variation of Functions. — A variable y is said to be a func- 
tion of a variable x when it depends upon x for its value, and so 
changes as x changes. It is often important to measure the rela- 
tive Tate of change of the function y with respect to its variable 



The Calculus. 




Fig. 2. 



x; that is, to tell, when a change is made in x, how many times 
as much change is made in y. In the case of an algebraic func- 
tion of the first degree, the relation is evident; for instance, if 
y = 2x + 7, an increase Ax given to x will change y to 2(x + Ax) 
+ 7, increasing it by Ay =2 Ax, twice. the increase in x. More 

generally, if y = mx + b, and x is 
increased any amount, y will be in- 
creased m times as much. This 
shows very clearly in the graph of 
the function — a straight line, where 
^ the increment given to x is repre- 
sented by the horizontal leg of a 
right triangle, and the correspond- 
ing increment of y by the vertical leg. All these triangles are 
similar for the same graph, and the ratio of the legs in each is 

_ JL = m. Thus from whatever value and by whatever amount x 

Ax J 

may be increased, y=mx + b is increased m times as much. 

When a function has a curved graph, the relation is more com- 
plicated; the ratio of the increment of y to the increment of x 
which produces it depends upon the initial value of x, the amount 
of the increment given to x, and the direction of the increment — 
whether it is an increase or a decrease. It is, however, of great 
value to extend the conception of the rate of increase of y rela- 
tively to x to mean the limit approached by the ratio of the in- 
crements of y and x as these increments approach zero together. 

^JL , is called the derivative of y with re- 

Az J Aaj=o 

spect to x. Before going more fully into the process of finding 
derivatives, we shall consider a problem illustrative of the mean- 
ing and use of the new idea. 

It will be understood, of course, that in this article the letters 
y and x have been used for function and variable merely for con- 



This limit, 



The Differential Calculus. 7 

venience and brevity; all sorts of letters may be expected in 
various problems. For instance, time is almost invariably indi- 
cated by the letter t, and distance (or space) by s. 

7. Speed. — The simplest idea of speed is familiar; if a train 
goes 80 miles in 2 hours, we say it has a speed of 40 miles an 
hour. The distance moved by any body divided by the time occu- 
pied in the motion is called the mean speed of the motion. 

The following notation for speed is used : 

80miles 4n » 80x5280 , Q2 » /m v 

-— ; = 40 m/n= s — ^ — =7; =o8*f/s(ft. a sec). 

2 hours ' 2x60x60 3 ' v ' 

When the mean speed of a body is the same for all intervals of 
time during the body's motion, regardless of when they begin or 
how long they last, the speed of the body is defined as being 
equal to its mean speed. When a train is stopped at a station 
or started from rest, its mean speed changes. If we compute 
the mean speed for an interval of time immediately following 
a given instant, then for shorter and shorter intervals also im- 
mediately following the given instant, we find that as the in- 
terval becomes shorter the mean speed changes, reaching a defi- 
nite value when the length of the interval is zero; this value is 
defined as the speed at the instant in question. 

Suppose, for instance, that a train is slowed down, and that 
it is known that the number of feet (s) run in t seconds after 
the power is shut off and the brakes are put on is given by the 
formula 

s = 60/-4/ 2 , 

what is the speed four seconds after, or when 2 = 4? When tf = 4, 

s = 60£-4/ 2 = 176; 

in the following table the computation of the mean speed is 
shown for several intervals immediately following the fourth 
second since the train began to slow down : 



The Calculus. 



t at end of 
interval. 


s at end of 
interval. 


Distance covered 
during interval. 


Length of 
interval. 


Mean speed 
duringinterval. 


5 sec. 

4.1 

4.01 

4.001 

4.000001 ° 


200. 

178.76 

176.2796 

176.027996 

176.000027999996 


24 ft. 
2.76 
0.2796 

0.027996 •' 
0.000027999996 V 


1 sec. 

0.1 

0.01 

0.001 

0.000001 •• 


24 f/s 

27.6 

27.96 

27.996 

27.999996 " 



This mean speed is very evidently approaching the limit 28 f/s. 
We can demonstrate this rigorously by finding the mean speed 
for any interval, At seconds long, immediately following the 
instant when t — 4:. Call the distance gone during the interval 
As (it is the increase in s caused by the increase At in t). 
When t = ± + At, s=176 + As; so 

176 + As=60(4 + A£)-4(4 + Ai0 2 = 176 + 28A*-4(A0 2 ft., 

and 

As=28A£-4(A0 2 ft. 

The mean speed during the interval of At seconds is : 

Mean speed = -^- = 28 - 4A£ f/s. 

If we replace At in this expression for the mean speed by the 
values used for the length of the intervals in the table, we obtain 
the numerical values given in the table for the mean speed ; and 
it is now evident that if we make At = 0, we get for the actual 
speed after four seconds 



ds 
~dt 



As 



At 



[28-4A*] At _ // JJ = 28f/fl. 



A*=0 



8. We can find the speed when the train has gone any num- 
ber, t, of seconds since the slowing down began. As before, take 
any interval At seconds just after the instant in question, and let 
As be the distance gone during the interval, s the distance gone 
at the beginning of the interval : 



The Differential Calculus. 



s=60t-4t 2 , 
s + As=60(* + A*) -4(* + A*) 2 ; 
As=60A*-82A*-4(A*) 2 ; 
60-8*-4A* 



As 
A*' 



is the mean speed during the interval ; the speed is 



As_ , 



ds 

dt i At J u =o 
The speed is zero when t = 7% sees.; then s = 225 ft. We thus 
see that after the brakes are put on, the train will go 225 ft, in 
7-J sees, and then stop. 




9. There is a very important connection between this problem 

and the preceding one, a relation which becomes apparent when 

we draw the graph of the function s = 60t — ±t 2 . In Fig. 3 this 

graph is drawn so that a unit length for s represents 50 ft., and 

a unit of t represents 1 sec. Each step in the process of finding 

the speed at the end of t seconds can be shown graphically: the 

point P shows a distance of s ft. traversed after t sees.; the 

point P' shows (s + As) ft, after (t + At) sees.; the mean speed 

As 
in this interval, or — , is the slope of the secant line PP' ; the 



10 The Calculus. 

limit approached by the mean speed, or the actual speed at P, 

is the slope of the tangent line at P. 

According to the notation of the preceding article, we should 

ds 
represent the slope of this graph by tan t— — , where dt is any 

convenient length (drawn to represent 3 sees, in the figure) and 
ds is so taken that -^ shall be equal to the value of 



At 



At=0 



For this reason we have used the fraction -57- to represent the 



Notice that tan t= 60 — St becomes zero when t — ^\, and is 
negative for larger values of t, as appears, either from inspection 
of the graph or of the algebraic expression 60 — St (r may be 
measured from the axis of abscissas to either part of the tangent, 
and changes from the third quadrant to the second, or from the 
first to the fourth, as t passes through the value 7-J). This cor- 
responds in the problem to a speed which decreases to nothing 
and then becomes negative. The problem as stated and the graph 
representing it end at the value t = 7-J, indicated by the point A ; 
if, however, the train were stopped by reversing the engines, it 
would be only momentarily stationary and then begin to back. 
In this case, the backing would be indicated by the negative 
speed, and represented by the part of the graph beyond A. 

Note again the difference in the use of the symbols com- 
pounded with A and those compounded with d. As and At rep- 
resent simultaneous variable increments, made to approach the 
limit zero (and reach it) ; ds and dt represent any fixed values 
whose ratio is this limit. Sometimes dt, ds, and As are grouped 
together as simultaneous increments; in this case all three rep- 
resent constants; dt represents some increment of time, As the 
actual distance traversed in this interval, and ds the distance 
that would have been traversed if the speed at the beginning of 
the interval had remained unchanged throughout the interval. 
On the graph (Fig. 4), dt appears as an increment of the abscissa 



The Differential Calculus. 



11 



of the point P, As as the corresponding increment of the ordinate 
of the curve, and ds as the corresponding increment of the 
ordinate of the tangent. 



s 










250 




P^L 


^r 




P' 


200 


as 

V 


ds -^ 
r 


/50 




eft 






100 










so 











4 5 6 

Fig. 4. 



/a 



10. Examples. 

1. A body falling freely near the earth describes a distance 
s = 16t 2 ft. (nearly), when t is the elapsed time in seconds; find 
the velocity at the end of 1 sec, at the end of 2 sees., and at the 
instant of starting. 

2. A body projected vertically upward from a height, ft, with 
a velocity, v, will at the end of t sees, have gone to a height 
s = li+vt— 16t 2 ft.; in each of the following three cases, find the 
speed after t sees., and the speed when the body reaches the 
ground; and in each case draw the graph representing s as a 
function of t, and draw t to represent speed = tan t. 

Case 1: ft = 100, v— +18 (bodv thrown upward at 18 f/s). 

Ans. 18-32* and -82 f/s. 
Case 2: 7i = 100, v—— 18 (body thrown downward at 18 f/s). 
Case3: ft = 100, v = (body dropped). 

11. The General Problem of the Derivative. — In the problem 
of the speed of the train, we have an instance of the general 
problem of the differential calculus ; the problem, that is, of find- 
ing a measure for the relative rate of increase of a function as 
compared with its independent variable; the distance traversed 



12 The Calculus. 

by the train is the function 60/ — It 2 of the number of seconds 
elapsed since the brakes were applied. We may have occasion to 
consider the rate of increase of any function relatively to its 
variable ; for functions of other variables than the time, this rate 
will have different meanings, but for any function whatever its 
measure is the slope of the tangent to the graph of the function, 
and the method of its determination is the same as for the time- 
rate or speed. 

If a point is moved along the graph of any function from left 
to right, it will go up or down according as the function in- 
creases or decreases with an increase of the argument. The 
curve shows by its steepness how fast the function changes in 
value as compared with its argument. The direction of the curve, 
which changes from point to point, is at any particular point the 
same as the direction of the tangent at that point, and is deter- 
mined by the angle r from the axis of x to the tangent. Thus the 
value of tan r for the graph is naturally the measure of the rate 
of change of the function, as compared with its argument. 

The rate of change of any function as compared with its argu- 
ment, or the derivative of the function, is always determined in 
general form, as a function of the argument; the determination 
for a particular value of the argument has been made in these 
early examples merely as a means of simplifying the conception 
of the subject. 

The distinctions already noted are made in general between 
the uses of the symbols compounded with A and those com- 
pounded with d; and, in addition, it is customary for the sake of 
brevity, when the function is indicated by y or by f(x), to indi- 
cate the derivative by y' or f(x) (read "y prime," "function 
prime x"). We thus have the general definition or notation of 
the following article. 

12. Derivative of a Function. — Given the function y = f(x), 
its relative rate of increase as compared with its argument x is 



The Differextial Calculus. 



13 



called its derivative with respect to the argument, and is deter- 
mined as follows: 



f(x) = 



f(x + Ax)- 
Ax 



«*) 



Az=0 



Ax 



dy 

dx 



The law according to which a change in x causes a change in 
y or f(x) is more explicitly stated by giving the value of the dif- 
ferential of the function : 

dy — f{x)- dx or df(x) =f(x)- dx. 

Finding either the derivative or the differential is called dif- 
ferentiating the function. As an example, we have already shown 
that if 

y=f( X )=X 2 , 



tf=f(x) = 



(x+Ax) 2 — x 2 
Ax 



-2x-ty- 

dx 



_ Aa<=0 

dy — 2x - dx; d(x 2 ) — 2x • dx. 

That is, the change in x 2 is (momentarily) 2x times the change 
in x. 

13. Relation between Derivative and Graph. — If the tangent 
to the graph of y=f(x) makes the angle r with the axis of x, 

2/'=f(z) = -§=tanr. 

When a function increases as its argument increases, its de- 
rivative is positive, and the angle r for its graph is in the first 
or third quadrant; when a function decreases as its argument 
increases, its derivative is negative, and the angle t is in the 
second or fourth quadrant; when, as the argument increases, the 
function is neither increasing nor decreasing, but changing 
from one state to the other, its derivative is zero and t = 0° or 
180° ; i. e., the graph is parallel to the axis of abscissas. The 
function in this last case is said to have an extreme value, or an 



14 The Calculus. 

extremum; a maximum or a minimum according as the value is 
greater or less than the adjoining values on either side. 

The reader should notice, in the examples already expounded 
and in those that follow, that the determination of the value 

— y~ = —f- —1/ in every case consists in evaluating an 

^ X J A#=0 dX 

indeterminate fraction, a function of Ax in the form -=r . (See 
Algebra, Art. 45.) 

14. Examples. 

Apply the formal definition of the derivative to the determina- 
tion of the following : 

1. Find the derivative of ax + b with respect to x. (a and b 
constants.) Ans. a. 

2. Find the ^-derivative of x 2 + a. Ans. 2x. 

3. Find the ^-derivative of bx 2 . Ans. 2bx. 

4. Find the differentials of x s and of at 3 . Ans. 3a?dx, 3at 2 dt. 

15. Rules for Differentiating. — The process of determining a 
derivative by means of the formal definition is used only to estab- 
lish a few general rules, from which we can find the derivative 
of any function or combination of functions. Several of these 
rules were established in the Alegbra (Arts. 86-87, 113-114). 
These are here repeated in the new notation, with some addi- 

d df(x) 

tions. The notation-^— (f(x)), for ' , v ' is merely a matter 

of convenience when the parenthesis is long. 

The derivative of the sum of two functions is the sum of 
their derivatives. If f(x) and <f>(x) are two functions of x, we 
have by definition 



£uw ++(*)] = 



f{x + Ax) + <f>(x + Ax) - [f(x) + <j>(x)] 

Ax 



Aa;=0 



The Differential Calculus. 15 



f(x + Ax)-f(x) <j>(x + Ax)-<}>(x) 

Ax Ax 



Ax=0 



d[f(x)+<f>(x)]=f(x)dx + cf>'(x)dx; 
so that the same rule holds for the differential of a sum. 

The derivative of the difference of two functions is the dif- 
ference of their derivatives. The proof is precisely similar. It 
is evident that these rules may be extended to apply to the deriva- 
tive or differential of the algebraic sum of any number of func- 
tions. 

The derivative of a constant is zero. This is obvious, as the 
change in a constant c, corresponding to the change in any 
variable, is nothing; or the change in the constant is zero times 
the change in the variable. Again, the graph of y — c is a 
straight line, having t = and tan t = for all values of x. 

As a corollary of this and the preceding, the derivative of 
[f(x) -\-c\, where c is any constant, is f(x), or 

-^[f(x) +*]-/'(*). 

Also, 

d[f(x) + c] = f'{x)dx. 

The derivative of the product of a constant and a function 
of x is equal to the product of the constant and the derivative of 
the function. 

For by definition, 

^ c[f(x + Ax)]-cf(xj 
Ax 



ii«wi= 



= cf'{x). 

Ax=Q 



As a differential, 

d[cf(x)] = cf'(x)dx. 



The last two rules are conveniently stated : In differentiation, 
a constant term vanishes; a constant factor persists. 



16 



The Calculus. 



-j-{xn)=nx"-i, or d(x") = nxn-i dx.— By definition, 
(x+Ax) n -x n 



Bv the binomial formula, 



Ax 



A:r=0 



(x+Ax) n = x n + nx n - 1 Ax + n ^ n ^ x n ~ 2 (Ax) 



+ terms containing higher powers of (Ax) 



Hence 



(x + Ax) n — x r 

Ax 



Ax=0 



71X 



n-i^n(n-l) xn _ 2Ax 



2 



+ (higher powers of Ax) 



nx r 



Az=0 



This theorem is here proved on the basis of the binomial 
theorem, itself proved (Algebra, Art. 54) only for positive in- 
tegral values of n. The extension of the proof to other values of 
n is made in a later article of this book (Art. 19.) 

Derivative of a Product. — The derivative (or differential) of 
the product of two functions is the sum of the products formed 
by multiplying each function by the derivative (or differential) 
of the other. 

—r- (uv) = v—i ^ u ~j-~! or d(uv) = vdu J rudv, where u and v 

are functions of x. — Let Au and Av be the increments of u and v 
corresponding to the increment Ax of the independent variable 
x; then, by definition, 



dx 



(vv) 



Uu 



+Au) (v+Av) — ttvl 



Ax 



jAar=( 



[ v • Au+ w Av + Au- Av ~\ 



The Differential Calculus. 17 

Aw . Av Aw Av 

[Awl _du TAvI dv j|~ Aw Av a "1 

SiL-0 " S ; L^L=o " Si and [.Si • Si •■ A *_Uo = °- 

Hence 

d , N du . dv 

_ (w)=l ,_ +tt _, 

or 

d(wv)=V' <?w+w- dv. 

This formula enables us to differentiate the product of any 
number of functions of a single variable; and in combination 
with other rules to differentiate any polynomial function of two 
or more variables. For example, 

d(x 3 -3x 2 y + y 2 -2xy + 3x) 

= 3x 2 dx-3[x 2 dy + yd(x 2 )]+2ydy-2(xdy + ydx)+3dx 
= (3x 2 -6xy-2y+3)dx- (3x 2 -2y+2x)dy. 

16. Examples. 

1. Find the ^-derivatives of the following functions: (a) 
2-3x + x\ (b) x 2 -x\ (c) x* -14z 2 + 24z + 12. 

Ans. (a) 3(z 2 -l); (b) z(3z-2); (c) 4(^-7a; + 6). 

2. Find the value of x for which each of the derivatives of 
Example 1 is zero. Ans. (a) ±1; (b) 0, f ; (c) 1, 2, — 3. 

3. Trace the curve y=x 2 — x s , and find the equation of the 
tangent at the points where the curve intersects the ar-axis. 

Ans. y=0 and x+y=l. 
Find the derivatives of the following : 

4. x\ lOz 4 . Ans. 7z 6 , 40z 3 . 

5. x 5 -3x s + l'7x. Ans. 5z 4 -9a: 2 + 17. 

6. at 2 — bt + c. Ans. 2at — b. 
Find the differentials of the following: 

7. -as 2 , 6s 6 . Ans. -2asds, 36s 5 <fe, 

8. — -V" + v £+s. Ans. ( — gt+v )dt. 

9. — 0# + v o . Ans. —gdt. 
3 



18 The Calculus. 

10. In each of the following identities;, find the derivative of 
the left member by the rule for the product, and check by find- 
ing the derivative of the right member. 

11. Differentiate u-u=x by the rule for the product, finding 

U ~W = ^ and thence snow tnat ~d£ ^2Vx' 

17. Derivative of an Implicit Function (Algebra, Art. 114). — 
When a function is defined by an implicit relation such as 
x 2 + y 2 = a 2 , or, in general, f(x, y) =0, it is understood that x 
being given any value, y must have such a value as to make the 
relation true, and is thus confined by the relation to a value or 
values depending upon the value of x; y is, in other words, a 
function of x in the ordinary sense. For any pair of values of 
x and y so related, then, x 2 .-\-y 2 or f {x, y) is a constant, and its 
differential is zero. Differentiating x 2 + y 2 = a 2 , we thus find that 

2xdx + 2ydy = 0, 
or 

dy _ _ x_ 
dx y ' 

thereby determining the derivative of y with respect to x. 

Whatever may be the form of f(x, y), the equation df(x,y)=0 
involves dx and dy, so that differentiating any implicit relation 
between two variables results in an equation involving their dif- 
ferentials, which can be solved to give a value of the derivative of 
one with respect to the other. 

18. Examples. 

Find the derivative of y with respect to x in each of the 
following : 

dy _ y — 2x 
1. x 2 -xy + y 2 = l. Ans. dx~2y-x' 



The Differential Calculus. 19 

2. fa'-8if+4*-«*+*=.0. Ans. | = |±|- 

3/^-3^+*-^. An, | = _-g5|±l. 

19. The Differential of jr" when /* is Fractional or Negative. — 

We have proved (Art. 15) that dx n = x n ~ 1 dx for the case in which 
n is positive and integral. Suppose now that n is a negative 
integer, and let n = — m, m being of course a positive integer. 
Let y=x n = x~ m ; then 

x m y=l. 

Then (Arts. 15 and 17) : 

x™ dy + mx m ~ 1 ydx = ; 

-£- — —mx~ x y — nx n ~ x . 

Thus the rule is proved for all integral values of n, positive or 
negative. 

Suppose, further, that n is fractional (either positive or nega- 
tive), and let n= — , p and q being integral. Let y = x n = x' P 9 ' ; 
then 

yQ = xP, 
qy q ~ 1 dy = px p ' 1 dx, 

dy__ p x*- 1 _ p .^jLfe-u p p=3 
ax q y q q q 

^- = nx n - 1 . 
ax 

This proves the rule for any rational value of n, positive or 
negative, integral or fractional; we are thus enabled to find the 
derivative of any algebraic expression. 



20 The Calculus. 

20. Examples. 

Find the derivative of each of the following: 

1. \=x~ 2 . Ans. -2x- 3 =^. 5. II— . Ans. -fart. 

x 2 x 3 y x 2 

2.—. Ans. -i. 6.-4=r. Ans. 



a? a; 2 ' ' vV 2V# 5 * 

3. VJ=s*. Ans. |^=^V. 7. *•-<*. Ans. f^-JK 

4. VF. Ans. fVz. 8. s 2 -V~s. Ans. 2s-— !-^ 
Find the differentials of each of the following : 

9. V^F. Ans. iVat'dt. U- 2 a/jL. Ans. -^=. 

Find the value of -^- from each of the following : 

»-(f)'+(i)'=>- *-*-;*■ 

21. The Differential of a Function of a Function. — If we are 

told that of three men, A, B, and C, A does twice as much work 
under given conditions as B, and B three times as much as C, 
we conclude readily that A does six times as much as C. By the 
same connection of ideas, if y is a function of x, and z is a func- 
tion of y; for instance, if y—x 2 and z = y 3 , so that the rate of 
increase of y is 2x times that of x, and the rate of increase of z 
is 3y 2 times that of y, the rate of increase of z is clearly 
3y 2 X 2x or 6xy 2 times that of x. This is formulated in general : 
If y—f{x) and z = <f>(y), so that 



The Differential Calculus. 21 

dz dy dz „, v , t/ * 

i& = iit-w =n * ) " , ' (y) - 

The identity is sufficiently evident from the fact that dy can- 
cels out from the differential expressions for the two derivatives. 
The importance of this relation in actual work is very great; 
there is indeed no single principle upon which so much depends 
in the practical use of the calculus. 

Suppose we wish d(x 2 -\-ay. We may say: 

Let y— (x? + a) ; then 

d(x*+a)*=d(tf)=Btfdt,j 

but dy—2xdx; so 

d{if) =oif • 2xdx = 10x(x 2 + a) 4 dx. 

With a little practice, however, most of these steps may be 
omitted or taken mentalh T , as follows : 

d(x 2 + a) 5 = 5(x 2 + a) 4 d(x 2 + a) 

= o(x 2 + a) 4 2xdx 

= 10x(x 2 + a) 4 dx. 

x dz 

Again, given x 2 +y 2 = ar, z — ; to find -j- as a function 

of x and y. We have : 

2xdx+2ydy=0; 

dy -_ x m 
dx y 

And again, 

z=-xy~ 1 , 
dz = xy~ 2 dy — if^dx, 
dz _x dy _! x I x \ 1 



_ y 

dx y 2 dx y 2 \ y I y ' 

dz x 2 1_ _ x 2 + y 2 

dx~ if " y "' y 3 

Since x 2 + y 2 = a 2 , this can be simplified: 



dx y 



22 The Calculus. 

22. To most persons the simplest way to use the rules of dif- 
ferentiation is to learn the expressions for fundamental differen- 
tials, and to regard the derivative of a f(x) as the result of find- 
ing the differential and dividing by dx. 

Moreover, as the principle mentioned in Art. 21 is of the 
greatest importance, it is well to memorize the fundamental 
rules in such a way that it shall not be readily possible to lose 
sight of its application. So stated, the rules applying to alge- 
braic functions are as collected below : 

If u and v are any functions of the independent variable, 

d(u±v)=du±dv, 
d(uv) =udv+vdu, 
d(u^) =nu n ~ 1 du, 

No other rules for algebraic functions should be memorized 
except as they force themselves on the attention through per- 
sistent occurrence; unless an exception be made in favor of the 
rule for a quotient, a special case of the rule for a product, which 
is readily seen to be 

-, u n -, vdu — udv 
d — = duv -1 = — —z . 

V V 2 

This rule is useful only when both terms of the fraction are 
variable. If either term is constant, the rule for u n is simpler to 
apply. 

23. Examples. 

1. Prove the rule for differentiating a quotient. 

2. Show that d ^y = x *_ Ux +± 9 dx. 

Show, without first expanding the expressions to be differ- 
entiated, that: 

3. d(a-x) 4 = -4:(a-x) 3 dx. 

4. d(3a + 2x) n = 2n(3a + 2x) n - 1 dx. 

5. d(4:-7x 2 ) 5 =-H0x(4-7x 2 ) 4 dx. 



The Differential Calculus. 23 

6. d(3-2x 3 ) 6 =-36x 2 (3-2x 3 ) 5 dx. 

7. d(a* + a 2 x 2 + x*) 3 = 6x(a 2 + 2x 2 ) (a* + a 2 x 2 + x*) 2 dx. 

8. d(a — x) 2 (a + x) 3 = (a — x) (a-\-x) 2 (a — bx)dx. 

n 7 a —adx , a adx 

9. d — — = t — ; — r 2 i d = 



x+a ~ (x+a) 2 ' a—x~ (a — x) 2 ' 

[Xote: Use u n rule.] 
10. d(2a-3x) 2 (3a-2x) 3 =-30(a-x)(2a-3x)(3a-2x) 2 dx. 
/j2 4:Q, 2 xdx 

n - d W+¥y = (a°+a?y ^ T ° te: Use u " rule -3 

12. dVa 2 + x 2 = , xd 2 X . [Note: =d(a? + x 2 )i.] 

V a i 2' 

13. d 



Va 2 -x 2 V(a 2 -x 2 ) 

/a — x _ 
V a+x ~ (a 



— aax 
. a 



(a + x)Va 2 -x 2 ' 



la+x _ 
V a—x ~ (a- 



adx 



x)\/a 2 — x* 



(a + x) 3 (a + x) 4 

(3a-2x) 3 _ 6(a + x)(3a-2x) 2 
17 • d (2a-3x) 2 ~ (2a-3x) 3 dx ' 

18. d(a-x)$(a+x)l= -$(a+2x) (a-x)-l(a + x)-*dx. 

_ _ 1 — dit . /- dw. , 1 — du 

19. d— = -zt~, dVu= ^—^:, a—;?. = -zr> 



w w z ' ~ 2%/u \/u \/u l 



A 1 -2du 

d — ; i = =— 



20. dV(a 2 + 3^) 3 = 9zVa 2 + 3z 2 dz. 

T g 2 + 2a&:r + & 2 :r 2 _ o(a + &x) 
* 1b d 2(a+6) 3 ~ (a + b) 3 dx ' 



Va—x—Va+ 



x 



22. d[Va-x+Va+x~\ = , , 2 — da 

L 2ya 2 —x 2 



i. ^ 



;2x 3 -3a 3 _ z 2 V3dz 



3a Va(2^-3a 3 ) ' 



24 The Calculus. 



, x 2 — a 2 4a 2 xdx 



J I' 
' \' x* + a 2 3(x 2 -a 2 )%(x 2 + a 2 )*' 

25. d i/ a ^4 = -a 2 x(a 2 -x 2 )-i(a 2 + x 2 )-Zdx. 

26. rf(w + i;) w = w.(^ + v) w - 1 (^+Ji;). 

27. d(u tl + v n )=nu n - 1 du+nv n - 1 dv. 

28. d(a+U n ) m = nmb(a+M n ) m -H n - 1 dt. 

M. d «_= ~*i 2 l p -\ C p dp. 
bp 2 — cp s (bp 2 — cp 3 ) 2 e 

30. If x=at % , y=bt, d(x 2 +y 2 ) =2t(2aV + b 2 )dt. 

31. Iix=at 2 ,y=U,d^ = =^. 

Z%:U X =at%y=U,d^ y = ^-. 

33. In 30-32, find the corresponding derivatives with respect 
to x (see Art, 21) : 

d(x 2 +y 2 ) = 2t(2a 2 t 2 + b 2 )dt _ 2a 2 t 2 + b 2 
dx 2atdt ' a 

dx\x)~ 2aH 3 ' dx \2ay) ~ 4a V " 

34. if y= ly?z?^ a -** 2 , 

^ a ' dz aVa?-x 2 

24. Trigonometric Functions. Circular Measure. — When we 
compare the increase of a function with that of its argument, 
the units used are of essential importance. For instance, in the 
speed problem of Art, 7, if the speed had been computed in 
yards a second rather than feet a second, the rates would have 
been one-third as great numerically ; if in yards an hour, twenty 
times as great. 



The Differential Calculus. 



25 



Any trigonometric function, such as sin 6, being the ratio of 
two lengths, is an abstract number, so that there is no question 
of its unit; but the angle itself may be measured in terms of 
various units. Of these the most convenient for use in the 
calculus is the radian, or the unit of circular measure; and in 
.this subject it is invariably the unit used. So it must be remem- 
bered that for the purposes of the calculus, the argument of any 
trigonometric function is always understood to be expressed in 
circular measure. 

As this understanding always exists, it is customary to use 
merely the name of the argument, without reference to the unit ; 
thus we speak of comparing the changes in sin and 6, meaning 
by the circular measure of 6, or the value of 6 in radians. 

When the name of an angle is used to indicate a quantity, the 
quantity indicated is always the circular measure of the angle. 

25. Derivatives of Trigonometric Functions. — In order to find 
the derivatives of sin 6 and cos 6, we shall need to know tho 
values when A0=O of certain forms. Construct a figure (Fig. 
5) showing graphically the circular 

measure I — J , the sine f — J and 

the cosine ( — J of an angle A0, 

and add the tangents as shown. 
Note that doubling A0 doubles also 
s, a, and t. As A0 approaches zero 
as a limit, c approaches r as a limit, 
and s, t, and a each approach 
zero. 




Fig. 5. 



By geometry, 

2s<2a<2t, 



r r 



or: 



sinA0<A0<tanA0, 
A0 



K 



sin A0 



<secA0. 



26 



The Calculus. 



Now 



hence 



sec A0= — 



J A0=O 



A0 



sin A0 



= 1. 



&e=o 



By means of this proposition, we can now find the derivatives 
of sin and cos 6. 
By definition, 



^sinfl . 
dO 

which may be written 

^sin 6 
dO = 



" sin(fl+Afl)-sinfl 

A0 A 0=O 



" 20+A0 . A0 

2 cos — - — sin —=- 

z z 

Ad Jaa= 

dsmO 



JA0 J ^=o 



dO 



= cos 0, 



since the limit of the fraction is 1, or 

(/(sin 0)= cos Odd. 
Similarly, 



dcos 



dcosO 



d$ 



" cos(fl + Afl)-cosfl 

A0 J A0=O 

sin#, 



or 



</(cos0) = -sin0(/0. 

We can find these derivatives much more simply by making 
use of the expressions for the sine and the cosine as algebraic 



The Differential Calculus. 27 

functions of the circular measure, which are shown to be (see 
Brown's Trigonometry, Art. 66). 

cos^l- ¥ + ¥ - 1+ ...., 

whence the derivatives of the sine and cosine of an angle with 
respect to its circular measure are 

<2sin0_, P 0* P 

Str - 1 - -jjr + "jT " w + ' * * ' * cos ' 

We can now differentiate the other trigonometric functions, 
which are algebraic functions of sine and cosine. 

iJ™l = jj f (cos 6 )--= - (cos ey> ( -sin $) 

sin A , „ 

= — 5-^ = sec tan 0. 
cos- 

i^* = -fL (gin*)-i=- (sin «)-*(«»«) 

COS /i x /) 

= ^— . = — csc cot 0. 

sin 2 6 

-^tam6=~ sin0(cos0)- 1 

= -sin(9(cos0)- 2 (-sin0)+(cos0)- 1 cos0 
= l+tan 2 0=sec 2 0. 

sin 9( — sin 0) —cos fl(cos 0) 
sin 2 

= -^-=-csc 2 0. 
sin 2 

One of the last two has been derived by the rule for the 
product, the other by the rule for the quotient; each is readily 
handled in both ways. 



d , a d /cos 6\ 



28 The Calculus. 

26. The principle of differentiating a function of a function 
by successive application of several rules (Art. 21) occurs con- 
stantly in treating trigonometric functions. E. g., 

d(sin 3 20) =3 sin 2 2(9 J(sin 20) 

= 3 sin 2 20 cos 20 d(20) =6 sin 2 20 cos 20 dO. 

It is likely to be the case that a trigonometric expression can 
be put in several different forms, one of which may be simplest 
for one purpose, another for another purpose. For instance, we 
might write 

3 sin 40 sin 20 dO for 6 sin 2 20 cos 20 dO, 
or 

24 sin 2 cos 2 0(cos 2 0-sin 2 0)d0, 

or might express this last form in terms of sin 2 only or of 
cos 2 only. The answers to the examples may need some such 
reduction before they agree with those in the text. 

27. Examples. 

1. Prove the rule for d tan from the rule for d—. 

v 

2. Prove the rule for d cot from the rule for duv. 

3. Since, if sin = 2/, and cos6=x, we have x 2 + y 2 = l, 

fill n T 

-~jn- =x > prove from these two relations that—^~= — y. Again, 

prove this from the relation cos = sin(£— 0). 
Prove the following : 

4. dsin 2 £ = sin 2x dx— — dcos 2 x. 

5. dsm(x 2 ) =2xcos(x 2 )dx. 

6. dcos(x 2 ) = — 2xsm(x 2 )dx. 

7. d Vl-cos = ±Y2 cosldO. 

8. JVl + cos (9= -jY2 sing dO. 

y i+cos0 2 

1U - a l + cOS0~ 1 + COS0* 

11. J(csc0-cot0)=isec 2 fd0. 



The Differential Calculus. 



29 



12. d cos 50 =— 5 sin 50 &0. 

13. d(sin 70-sin 30) = (7 cos 70-3 cos 30)d0. 

14. d cos 50 sin 20 = |(7 cos 70-3 cos 30)d0. 
2 sin t 



15. J tan 2 1 



dt. 



cos* 



16. <2 sin ax 2 — 2ax cos ax 2 dx. 

17. If £ = <zcos<£ and ?/ = 6 sin </>, 

18. If x=asec<£ and ?/ = &tan<£, 



dy 
dx 

dy = & 



cot <f>. 

a 



esc <£. 






80 - Ia ( 18 >^(^)=-^- cot ^ 



CSC 3 <£. 



Compare Exs. 17-20 with Exs. 34-37, Art. 23. 

21. d(sec + tan 0) n -n sec 0(sec + tan 0) n . 

22. d(cos 4 0-sin 4 0) = -2 sin 2(9 d0. 

23. For what value of the angle is it's sine increasing one-half 
as fast as the angle ? The tangent twice as fast ? 

Ans. £ and J. 

28. The Inverse of the Trigonometric Functions. — The sym- 
bol sin' 1 (which is read "inverse sine") stands for "the angle 
whose sine is .... " When it is used in the calculus to repre- 
sent a quantity, it signifies " the cir- 
cular measure of the angle whose 
sine is .... " So for the other in- 
verse functions. In the familiar 
graphic representation of Fig. 6, the 
straight lines s, c, t, and the arc 
represent the sine, cosine, tangent, 
and circular measure of the angle 0. 
Thus s is the sine of the arc 0, and 
is the arc of the sine s. Conse- 
quently many mathematicians read 
the symbol sin~ x " arc-sine "; e. g., = sin _1 s = cos _1 c = tan _1 t is 
read : " equals arc-sine of s, arc-cosine of c, arc-tangent of t." 




Fig. 6. 



30 



The Calculus. 



29. Derivatives of the Inverse Trigonometric Functions. — The 

x or 

relations 0=sin _1 — . and sin0 = — < are equivalent; differen- 
ce a u 

tiating the second, we find that 



dx 



dx 



cos0d6=—- • d0 = 

a ' a cos 



dx 



hence 



VaF=x~ 2 ' 



dsm- 1 -^- = 



dx 



By the same method we derive the following list of differentials : 

</sin-i — = -, dx = -</cos-i — , 
a \/a2 — x2 a 

adx 



</tan-i — = 
a 



a * +jfi =-</cot-i T 



dsw -l± = f* 

a xVx 2 — a* 



= — (/C8C-1 



To these may be added, if versin 0=1 — cos 0, 

j i x dx 

(/versin-i — = —====. 
a \/2ax—x2 

It is helpful to notice, in order to remember when the factor 

x 
a appears in the differential of an inverse function of — , that 

such a differential is always of degree zero, a, x, and dx each 
representing a length, and so being of degree 1. 




^oP 



Fig. 7. 



The Differential Calculus. 31 

It should also be noticed that an angle may be given much 
more simply as the inverse of one function than as the inverse of 
another; the best method of transformation is to draw the geo- 
metric figure. For instance, 



a a 



x ci 



30. Examples. 

Prove the following : 
„ 7 . i x — 2 dx 

b—x dx 

2. d cos -1 



3. dtan" 1 

4. dtan- 1 



a Va 2 -b 2 + 2bx-x 2 

2x— 1 4dr 

2 ~ 4a; 2 -4a- + 5 * 
a: — b 2adx 



2a x 2 -2bx + b 2 + ±a 2 



5. -=— sin -1 (cos x) = — 1. 

6. -^ ■*<«** *)=r 7 =2 ? 



7. — =— (x sin -1 #+ Vl — # 2 ) = sin -1 # . 

dx 

8. -j— tan -1 - 



dx Va 2 -x 2 Va 2 -x' 

9. -T- esc" 1 ' — = , , 2 . 

ax x a 2 + x 2 

d n 

10. -j— sec 



dx y~aF^tf~ Va 2 -x 2 



^-S^ViTi — 57T 



or 
( Note : Let a; = cos 9 and write the radical as a function of f . ) 



32 The Calculus. 



12. 4- m 



s- 



—x 



13. 



dx™ VT" 2VT: 

d n . /1+x 1 



2Vl 



14. -A. ^-i fc, VI "* 2 =y= 

15. -4- ten- 



ds 1-x 2 1 + x 2 ' 

31. Exponential and Logarithmic Functions. — The derivatives 
of exponential expressions like a x , e x , can best be derived from 
their expansion into infinite series (see Algebra, Brown and 
Capron, Art. 115). The value of e x , thus expressed, is 

/m2 /v.3 n, 4 

'"=i+*+f +f + f + ••••' 

the derivative of which is 

dx ~ 1+ * + l!f + !- + ]T + - •••-'■ 
Hence the derivative of e x is e^ itself. 

de x = e* • dx. 

Differential formulas of logarithmic expressions are simplest 
when the base is e; logarithms to the base e (natural logarithms) 
are therefore used more than any others in the calculus; and so 
for convenience we write log x for log e x, and always understand 
the base to be e when no base is written. 

The value of dlogx may be obtained from the equation 
y=log e x, which in exponential form, is 

e y —x. 
Differentiating, we find 



so that 



#dy=dx. -^= * -1 
v dx . e 9 x 



, 4 dx 

dlogx=—. 






The Differential Calculus. 33 

The differential of a x may be obtained from the series for of, 
or as follows: liy — a x , 

logy=(loga)x, 

-^-=(\oga)dx } 

dy= (log a)ydx, 
d(a*) = a* « log a • dx. 

For d logo x we have, if y = log a x, 

x-ay } 
dx—a v loga dy, 

dx dx 



dy 



a y log a x log a 

As i = log a e, the modulus of the system having a as its 

iog e a 

base, 

</log ff jr=log a e — • 

When a— 10, the modulus, log 10 e = . 43429 + , is usually de- 
noted by ix, so that 

udx .43429(/jt 
rflog loJ r=^- = —j—. 

32. The Logarithms of the Trigonometric Functions. — By 

combining the principles of Arts. 25 and 31, we readily obtain 

the differentials of the logarithms of the trigonometric functions. 

For instance, 

, . d sin 6 cos dO , nAn 

d log sin 0= -^y = -^0- =cot M, etc. 

The full list is : 

</logsin0= cotO. dO, 

(/log cos 0= — tan0. </0, 

(/log tan 0= 2csc20.</0, 

</logcsc 0=_COt0. (/0, 

d log sec - tan . </0, 

4 tflogcot 0=-2csc20.</0. 



34 The Calculus. 

33. Logarithmic Differentiation. — It often happens that the 
logarithm of a function is easier to differentiate than the func- 

/ d 2 -i-x 2 

tion itself. For instance, finding d ju — ^ involves a good deal 

* (X X 

of algebraic work, which is lessened appreciably by the following 
process : 
Let 



V a 2 -x 2 > 



y- 

then 

log y=i log(a 2 + x 2 ) — \ \og(a 2 -x 2 ), 
dy xdx xdx 2a 2 xdx 



y a 2 + x 2 a 2 — x 2 a^ — x* 9 
i _ 2a 2 xdx _ J a 2 + x 2 2a 2 xdx 2a 2 xdx 



1 ~ (a 2 -x 2 )$(a 2 + x 2 )l' 

Besides being convenient for products, quotients, powers, and 
roots, logarithmic differentiation is necessary for exponential 
functions in which both base and exponent are variable. 

For instance, to find dx x ; if y—x x } 
\ogy=x\ogx, 

<-£- = dx-\- log xdx, 

dy=y(l+logx)dx=x x (l + \ogx)dx. 

34. Examples. 

_, d , x — a 2a 

1- "XT lo S 



dx x+a x 2 

2. A log*- 



dx x — b (x — a)(x—b) 

3 _^- e a sin x = ae a 8in x cos x. 
dx 

e x "* 

4 - y= l °Snr7x- y = 



l + e x a l + e x 



The Differential Calculus. 35 



5. y=\og(x+Vl + x 2 ). y'= / -_ . 

Vl + z* 

6.f(x)=hg(hgx). f{ x ) = -±-. 

8. /(0) =log(sec 0+tan 0). /' (6) = sec 0. 

10. /(0)=logtan(-+J). f(0)=sec0. 

V V(l-s) U {l-x)V{l-x 2 ) 
V(^+l)(^ + 3) 9 , z 2 Qr + 3)S 

La - V- {x~+zy y ~ (x+2) 5 (x+i)f 

35. "We shall proceed after this article to some applications of 
derivatives, and for some time shall develop no further rules for 
differentiating; it will therefore be convenient to collect at this 
place the rules we have so far derived. These rules are given as 
expressions for differentials. As a practical rule for finding 
derivatives, we have : 

To find the derivative of a variable z with respect to a variable 

dz • 

w, find the quotient—* — , and if any derivative appears in this 

quotient, determine and substitute its value. 

d constant =0. d esc u= — esc u cot udu. 

*d(u±v)=du±dv. . du _ . , 

*duv = udv + vdu. d sm_1 u = VT^ = ~ d C ° S U ' 

d u^vdu-udv^ dt^u = -^- 2 = -dcot->u. 

V v 2 1 + u 2 

*du n = nu n - 1 du. du -, 

a sec -1 u— — i 9 H = —a esc* 1 u* 
d l -du u\/u 2 -l- 

u ' u 2 ' de u = e u du. 



36 



The Calculus. 



■,, r- du 

*dsm u = cos udu. 
J tan u= sec 2 udu. 
dsecu=secu tan udu. 
d cos u= — sin udu, 
d cot u— — esc 2 udu. 



*da u = a u loga du. 
du 



dlog u— 



u 



dlog a u=log a e 



du 



dlog 10 u 



fidu .43429 du 



du? is found by logarithmic differentiation. 

From the rules marked with an asterisk, all the others can be 
derived. 

36. Miscellaneous Examples. 

Prove the following: 

1. -J- (0-sin0cos0)=2sin 2 <9. 



2. 
3. 



smx 



a cos x- 



dx a—b cosx ~ (a—b cosx) 2 ' 
d 



dx 



( 2x sin x + [2 — x 2 ~\ cos x) — x 2 sin x. 



4. -j- sin -1 
ax 



x 
1+a; 



-1 



l+x) Vx 



5 -^ Bin-* A / * 2 - a2 = zVa 2 -^ 
* dx * V z 2 -& 2 (x 2 -6 2 )V^ 

fi A . , g \/3 __ V3 

°- da; tan x + 2 2(x 2 + x + l) # 



7 * daT cos 



b + a cosx y a 2_j,2 



A 

da; 



a + frcosa; a+fccosa; 



'VoT- ] 



Va 2 -& 2 



a + & cos a; 

Solve the following by letting a; = asin<k finding the deriva- 
tive with respect to <f>, and thence the derivative with respect to x. 
d x a 2 



9. 



dx ^/a^~x~ 2 ~ (a 2 -x 2 )*' 



10. 



The Differential Calculus. 37 

_d_ y a 2 - x 2 _ 1 

dx a 2 x ~ x 2 \ / a 2 -x 2 ' 



Make similar substitutions in the following : 

11 JL x * - 3x 2 (l + x 2 ) 
' dx (1-x 2 ) 3 ~ {i-x 2 y ' 

-t q 0/ X oX 

* ~dx~ (1-x 2 )* = (1-x 2 )* * 

* dx VT+^~ (1 + z 2 )*' 

14. d log (sec 0+tan 0) =sec $ d$. 

15. d[sec 6 tan + log(sec + tan 0) ] =2 sec 3 &0. 

16. de-* 2 =-2:re-* 2 da;. 

17. dx sinx = x sinx ~ 1 smx+x sinx co&xlogx. 

37. Problems in Speed and Time-Rates. — Speed has already 
been defined as the relative rate of increase in the distance 5 
traversed from a fixed point, as compared with the time t elapsed 
since a given instant. It is equal to 

ds _ I* As" 

dt ''- L"A^JAt=o , 
the derivative of s with respect to t. Speed may thus be called 
the time-rate of distance. Any other variable has in the same 
sense a time-rate. Thus if 6 is the angle generated by the spoke 

of a wheel during any time, t, -rr is the time-rate of the rotation, 
called the angular velocity. Again if A is the area of an expand- 
ing surface, or V the volume of an expanding solid, -rr or --=- is 

the time-rate of expansion, and so on. All these time-rates are 
sometimes called velocities, or simply rates. They are all, of 
course, derivatives with respect to time. 

Aside from certain general principles., there is no theory of 
this application of derivatives; such difficulties as arise in the 



38 



The Calculus. 



solution of the problem are chiefly matters of algebra and 
geometry. 

We shall discuss a few examples suggestive of the methods 
most commonly useful. 

Example 1 : A rope attached to a boat is being hauled in at 
the rate of 2^ f/s by a man on a wharf, whose hands are 12 ft. 
higher up than the point of attachment of the rope. Find the 
speed of the boat (a) in general, (b) when it is 9 ft. from the 
wharf. 

Let y be the distance of the boat from the 
wharf at any time, and x its distance from the 
man ; then we have given 

and are to find -&. . Then 
at 




y=\/x 2 — 14:4:; dy — 



xdx 



Dividing by dt, 

dy x 

dt 



dx 



Vz 2 -144* 
— 5x 



V^-144 dt 



2 x /r 5 -144 
When y = 9, x= Vl44 + 2/ 2 = 15, 



f/s, in general. 



dy 
dt 



75 



Jy=9 



18 



f/s =-4i f/s. 



Note the principle actually employed here : The time-rate of 

x 
x is given = !• f/s; we find the anrate of y= , 2 =- ; 



then 



since the increase in y is 



Vz 2 -144 



times that in x, and the 



increase in x is --£- that in t, the increase in y is 



dy 



dy dx 
dx dt 



— 5x 



2V^-144 



times that in t. 



The Differential Calculus. 



39 



dx du 

Note that both -r- and -jr-are negative, since x and y both 

decrease. 

Note particularly that nothing can be accomplished by taking 
y = 9 at the start; the general case must be used to get an equa- 
tion expressing the functional relation between y and x. 

Example 2: Two railroad tracks 
cross at right-angles ; on each a train 
is approaching the crossing; one, 17 ** A 

mi. off, is going west at 12 m/h ; the 
other, 22 mi. off, is going south at 
15 m/h. Find the rate at which 
they are approaching each other (a) 

at the end of 40 min., (b) when the 

west-bound train is \\ mi. west of 

the crossing. After t hours, let the p^ . 

train going west be x mi. east of the 

crossing; the train going south, y mi. north of it; and let the two 

trains be z mi. apart. 



17 (1) 



We have : 



dx 



dy 

~dt 



12 m/h, -^- = - 15 m/h, 



z 2 = x 2 + - 



so that 



2zdz- 



dz 1 / 

dt ~ z \ 



2xdx + 2ydy, 

dx 
dt 



+y 



dy 
~dt 



)■ 



-J = 4" (-12z-15i/)m/h: 



^-{±x + 5y) m/h. 



At the end of 40 min., the trains will have gone 8 mi. west and 
10 mi. south respectively, so that x =9 mi., y = 12 mi., and 
therefore 2 = 15 mi. Hence, 



40 



The Calculus. 



dz 
~dt 



t=§ 



-A.(4X 9 + 5 X 12)m/h= — 9 /m/h. 



At the end of 40 min. the trains are approaching at 19.2 m/h. 

When x— — f, the west-bound train will have gone ^- mi. in -f-J 
hrs., and the south-bound train will be f mi. south of the cross- 



ing, so that y= — -f; consequently z 



-§-• 



Hence, 



dz 



dt 



t=u 



-3X8 
15 



[-4X|-5Xf]m/h = - 9 /m/h, 



and the trains are receding at 18.6 m/h. 

Since a: = 17 — 12t and y = 22 — 15t, it would be possible to ex- 



press z directly as a function of t, and thence find 



dz 



i, , but the 

computation would be much more laborious. The general results 
would be 

z=V773-1068£ + 369£ 2 , 
dz . -534 + 369* 

dt ~ V773-1068£ + 369£ 2 ' 

Example 3: A man is trotting around a circular track 4 mi. 
in diameter at the rate of 6 mi. an hour. Find the rate at which 
his distance from a fixed point of the track is increasing. 

Let the fixed point be A, and the man's position at any time 
be M; let the central angle AOM be $, and the straight line 
AM—x; then 

x=2 sin | (2 mi.) =4 sin | mi. 
dx=2 cos | dO mi. 




and 



dx 



= 2 cos | 



de 



m/h; 



Fig. 10. dt "~ % dt 

but in this relation 6 must be in circular measure. (See Art 24.) 



The Differential Calculus. 41 

As the radius is 2 mi., 6 mi. = 3 radii, and the speed of 6 m/h 
gives an angular velocity, ~-rr = 3 ; hence 

-^- = 3(2 cos | m/h) = 6 cos f m/h. 

With between and ir, this rate is positive, and the man is 
Teceding from A; with between ?r and 2?r, it is negative, and the 
man is approaching A ; this cycle then repeats. 

This solution can also be applied to any case of two objects 
moving in the same circle, their relative speed taking the place 
of the speed of the man in this problem. 

Examples. 

4. One end of a ladder 29 ft. long is against a vertical wall, 
the other on a horizontal floor. If the lower end slides along 
the floor at 1-J f/s, how fast is the upper end slipping down the 
wall when the lower end is 20 ft. from the wall? 25 ft. ? 

Ans. If ft. a sec; ff V6 f/s = 2.55 f/s approx. 

5. The free end of a ball of string is attached to a wall ; 8 ft. 
lower down, the ball is moving horizontally at the rate of 4 J f/s; 
how fast is the string unwinding when the ball is 15 ft. from the 
wall? Ans. 3f f/s. 

6. A stone drops from a height of 100 ft. upon level ground. 
Given that its speed when it has fallen s ft. is SVs f/s, find the 
speed of its shadow on the ground, when the stone is at a height 
of 9 J ft, the altitude of the sun being 30°. 

Ans. 76 V3 f/s = 131.6 f/s. 

7. If a shadow of the stone in example 6 is cast by a light 20 
ft. above the ground and 10 ft. from the path of the stone, how 
fast will this shadow be moving when the stone is 19 ft. above 
the ground ? 9f ft. ? Ans. 14400 f/s and nearly 144.7 f/s. 

8. A stone cast into a pool causes a circular wave, the front of 
which moves at the rate of 2^ in. a second. Show that the area 
of the circle increases at the rate of h-n-r sq. in. a second, r being 
the variable radius of the circle. 

9. A gun is fired from a balloon 200 ft. above the ground, 
producing a spherical wave in the air, which moves at the rate of 



42 The Calculus. 

1100 ft. a second. Find the rate at which the surface of this 
wave and the volume enclosed by it are increasing when the wave 
reaches the ground. 

Ans. Surface, at 1,760,000tt sq. ft./sec. ; volume, at 176,- 
000,000tt cu. ft./sec. 

10. Gas is pumped at the rate of 10 cu. in./sec. into a spher- 
ical toy balloon; how fast are the surface and radius of the bal- 
loon increasing: (a) when the radius is 5 in.? (b) When the 
balloon holds a cubic foot ? 

Ans. (a) Surface at 4 sq. in./sec; radius at -^ in./sec. (b) 
Surface at *£■ (J)* sq. in./sec; radius at -^ (£)* in./sec 

11. Two ships, A and B, are on the same meridian, 117 mi. 
apart. A is sailing due east at the rate of 10 m/h, B is sailing 
due north, toward A, at the rate of 15 m/h. At what rate are 
they approaching each other after they have been sailing 3 hrs. ? 
After they have been sailing 5 hrs. and 12 min. ? When are 
they neither approaching nor receding? What is the closest 
approach they make to each other ? 

Ans. +10 m/h, +1 m/h, after 5 hrs. 24 min., 18V13 mi. 

12. Two trains are 12 mi. and 6 mi. respectively from a cross- 
ing where their routes intersect at right angles ; the first train is 
approaching the crossing at the rate of 42 m/h, the second reced- 
ing from it at the rate of 36 m/h; are the trains approaching or 
receding, and at what rate: (a) after 10 min.? (b) After 2 
hrs.? (c) When will the distance between them be least? 

Ans. (a) Keceding, 17 T 1 - g -m/h; (b) receding, about 54.94 
m/h; (c) at the end of 5yymin. 

13. Two railroad tracks make an angle AOB = 60°; AO = S 
mi., BO = 5 mi. A train at A is approaching O at 20 m/h, and 
a train at B is approaching O at 30 m/h. Find (a) at what rate 
the trains are approaching each other; at what rate they will be 
receding or approaching: (b) when B reaches O; (c) when A 
reaches O; (d) at the end of 30 min. 

Ans. (a) 20 m/h; (b) approaching at 5 m/h; (c) receding 
at 20 m/h; (d) receding at about 22.9 m/h. 

rjop 

14. A solid expands so that the time-rate -=■:- of the expansion 

of any one of its linear dimensions is Tex, Tc being a constant. 

Show that the time-rate -77 of its volume is 3TeV. 
at 



The Differential Calculus. 43 

15. The roadway of a bridge is 20 yds. above the roadway 
below, and the two run perpendicular to each other. One man 
is going over the bridge at 3 m/h, and another, directly under 
him, is going at 8 m/h. At what rate will they be receding from 
each other at the end of 3 min. ? Ans. About 8.54 m/h. 

16. A city street has a vertical wall on one side, and 75 ft. 
from it, on the other side, is a light. A man starts 15 ft. directly 
up the street from this light and crosses straight over at 4 m/h. 
Find the rate at which his shadow is moving horizontally along 
the wall (a) when he has gone 10 ft.; (b) when he has gone 
50 ft. Ans. (a) 45 m/h; (b) 1.8 m/h. 

17. Wine is poured into a conical glass 3 in. deep at a uniform 
rate, filling the glass in 8 sees. At what rate is the surface rising 
at the end of 1 sec. ? When it reaches the brim ? 

Ans. -J im/sec. ; J in./sec. 

18. A beam 20 ft. long rests against a vertical wall and a hori- 
zontal floor ; a bar is attached by hinges to its middle and against 
the angle of wall and floor. The beam slides down so that the 
angle 9 between this bar and the wall increases at the rate of 18° 
a second. Find the rate at which each end of the beam is mov- 
ing when = 30°, 45°, 60°. 

Ans. 0=30°, 45°, 60°; rate of upper end=7rf/s, tt V2 f /s, 
ttV3 f/s ; rate of lower end = 7rV3 f/s, ttV2 f/s, tt f/s. 

19. The connecting rod, PA, of a stationary engine is 5 ft. 
long; and the crank AC is 1 ft.; if the crank revolves at the 
uniform rate of two revolutions a second, find the speed of the 
piston-rod (or of the point P) when the angle PC A is 0°, 45°, 
90% 135°, 180°, 225°, 270°, 315°. Ans. If PC = x, when 

0=0°, 45°, 90°, 135°, 180°, 225°, 270°, 315°, 360°,-^- =0 

=^7rx/2, -4*, -PW2, 0, ^W2, 4rr, ^W2, 0. 

20. A wheel 3 ft. in diameter rolls along level ground at the 
uniform rate of 10 m/h. Find the rate of the forward motion 
of the bottom, top, foremost, and hindmost points of the rim. 

Ans. x being the horizontal distance moved by a point fixed 
on the rim, measured from the point where it touched the road, 
£ = .|(0 — sin0)ft, being the angle generated by the radius to 

this point, and when 0=0° 90°, 180°, 270°,-^ =0, 10 m/h, 



4-4 The Calculus. 

20 m/h, 10 m/h, the horizontal speeds of the bottom, hindmost, 
top, and foremost points respectively. 

21. A line tangent to a circle of 10-in. radius moves across 
the circle at the rate of -J in./sec, keeping always parallel to its 
first position. Find the rate of increase of the area of the seg- 
ment next the point of tangency when the line has moved J way, 
\ way, and f way across the circle. 

Ans. 5V% 10, and 5 VF sq. in./sec. respectively. 

22. A plane moves across a sphere of 10-in. radius as the line 
moved across the circle in example 21; find the rate of increase 
of the corresponding volume. 

Ans. 3 7^7r cu. in./sec, 50?r cu. in./sec., and 3 7j7r cu. in./sec 
respectively. 

23. An eccentric circular cam of radius a inches revolves about 
a point at a distance b from its center C; the point being 
in line with a rod which bears upon the rim. Let the rod bear 
upon the point R of the rim, and call OR = r, and the angle 
COR = 6. Show from the triangle COR that 

r — b cos 6+\/a 2 — b 2 sin 2 0, 

so that if the cam makes n revolutions a second, the speed of the 

,, ,. . dr ~ 7 T • /i , &sin#cos# 

rod s motion is -=7- = — airnb sin v -) — . 19 . == • 
at va 2 — b 2 sm~6 



CHAPTER II. 
Analytic Geometry. 

38. Geometrical Applications of Analysis. — The relation be- 
tween a function and its graph is utilized in two ways : first, the 
properties of a function may be made evident by reference to the 
graph; and second, the properties of a geometric locus may be 
studied by means of its equation. The second of these was ex- 
pounded to a certain extent in Chapter VII, Loci of Equations, 
of the Algebra. We shall begin here with a recapitulation of the 
results of that chapter. 

39. Relation between a Locus and its Equation. — If the values 
of the coordinates x and y are varied independently, the point (x, y) 
will move all over the plane ; but if x and y are dependent, owing 
to the existence of a relation f(x, y) = 0, the motion of the point 
(x, y) is restricted to a curve. If f(x, y) —0, when solved, gives 
y = F(x), this curve is the graph of F(x), and: 

The curve is the locus of f(x,y)=0; 

f{x, y)=0 is the equation of the curve; 

The coordinates of any point of the curve satisfy f(x,y)=0 
and the coordinates of any other point do not; or 

Any point whose coordinates satisfy f(x,y)=0 lies on the 
curve, and any point whose coordinates do not satisfy f{x,y)—0 
does not lie on the curve. (Algebra, Art. 98.) 

40. The Linear Equation and the Straight Line. — If the equa- 
tion f(x, y) = of a curve is of the first degree, or linear, that is, 
of the form Ax+ By + C = 0, its locus is a straight line. 

The constants of the line being the ^-intercept a, the ?/-inj;er- 

cept o, the inclination to the axis of abscissas r, and the slope m, 

, A C 7 C 

ra = tanr= ^- , a— -j- , b = — ^ • 



46 



The Calculus. 



A straight line is traced by determining two of its points, or 
one point and its slope. 

The equation of the straight line through the point (x x ,y x ), 
and having the slope m or the inclination r ± to the x-axis, is 

y-y 1 = m ( X - Xl ), 

or 

y-y 1 = tanT 1 (x-x 1 ). 

The equation of the straight line through any two points 
(*i>yx) and (z 2 >y 2 ) is 



V2 






A 1 u. 

a^'b 



In terms of the constants m> a, and b defined above, the equa- 
tion of any straight line may be written 

y — mx + b ( slope and intercept equation ) , 
= 1 (intercept equation). 

(Algebra, Arts. 99-101.) 

41. The Angle between Two Straight Lines. — It is evident 
from Fig. 11 that if two straight lines, (1) y = m 1 x+b 1 and (2) 
y=m 2 x+b 2 , are inclined to the axis of x at the angles t x and t 2 
respectively, the angle a between them is 




a = r<> — r-i 



whence 



, tan t 9 — tan r n 

tan a— 2 - — ±- . 

1 + tan t 1 tan t 2 

As the slopes of the lines 
are m 1 = tan r 19 m 2 = tan t 2 , 

, m 2 — m, 
tana=^ . 

l + m 1 m 2 

If the lines are parallel, a=0, tana=0, and m 1 = m 2 ; or the 
slopes are equal. 



Pig. 11. 






Analytic Geometry. 47 

If the lines are perpendicular, a =-^-, tana=co, and m x m 2 
= — 1, or the slopes are negative reciprocals. 

42. The Mid-Point and the Distance between Two Points. — 

The point half-way between {x x ,y x ) and (x 2 ,y 2 ) is 

(x 1 ±x 2 y 1 + y 2 \ 
[~2~> 2 )' 

The distance between these points is 

d=V(x 2 -x 1 ) 2 +(y 2 -y l ) 2 . 
(Algebra, Arts. 104-105.) 

43. The Distance of a Point from a line. — The distance from 
(x ly y x ) \oAx+By+C = Q is (Algebra, Art. 106) 

Ax^ + Bi^ + C 
VA 2 +£ 2 * 

44. Intersections. — The coordinates of the point or points of 
intersection of any two curves are the simultaneous solutions of 
the equations of the curves. (Algebra, Arts. 107-108.) 

45. Finding the Equation of a Geometric Locus. — To deter- 
mine the algebraic equation of the curve traced by a point which 
moves under given restrictions: 

(1) Eefer the problem to a pair of axes, letting the variable 
coordinates (x, y) be the distances from these axes of any point 
whatever on the curve; 

(2) State in the form of an equation in terms of x and y the 
definition of the curve or the conditions of the problem. (Algebra, 
Art. 109.) 

46. The Equation of the Circle. — The equation of the circle 
having a radius of length a and the point ( o, c) for center is 

(x-b) 2 +(y-c) 2 = a 2 . 
If the center is at the origin, the equation is 

x*+y*=a 2 . 



48 The Calculus. 



Any equation of the form ax 2 + ay 2 + dx-\- ey +f=0 represents 

a circle, the radius bein^ J + e ~ ' and the center at 

V 4cr 

("if' ir)' (Algebra, Art. 110.) 

47. Tangents to Circles. — For any point on the circle 

x 2 + y 2 = a 2 
we have 

2xdx + 2ydy = 0, 
dy __ —a; 
dx ~ y 

Then, if m 1 =tanT 1 is the slope of the tangent at the point 
0*w yi) (see Art. 4), 

x x 

and the equation of the tangent is (Algebra, Arts. 115-116) 

x x x + y x y = a 2 . 

In the same way, the tangent to (x— b) 2 + (y— c) 2 = a 2 is 

(x-x ± ) (x x -b) + (y-y x ) (y 1 -c) = 0, 
or 

{x x -b) (x-b) + {y x -c){y-c) =a 2 . 

48. Tangent to Circle in Terms of its Slope. — If a line having 
the slope m is tangent to the circle x 2 + y 2 = a 2 , its equation is 



y = mx±aVl + m 2 . 

If this line is tangent to any circle (x—b) 2 +(y—c) 2 = a 2 , 
its equation is 

y — c=m(x — b) ±aVT+m 2 . 

These forms of the equation are useful in determining the tan- 
gents' to a circle from an outside point. (Algebra, Arts. 117- 
118.) 






Analytic Geometry. 



49 



49. Loci of Quadratic Equations. — Any quadratic equation, of 
the general form 

ax 2 + bxy\- cy 2 + dx + ey+f=0 
represents some one of four types of loci : 

(1) Two straight lines (distinct or coincident, parallel, or 
intersecting). 

(2) A closed oval called an ellipse (of which the circle is a 
special case) : condition b 2 — 4&c<0. 

(3) A single-branched open curve called a parabola: condition 
b 2 -4ac=0. 

(4) A two-branched open curve called a hyperbola: condition 
5 2 -4ac>0. 

In the Algebra, the circle was defined as a geometric locus, 
and its equation found ; but the ellipse, parabola, and hyperbola 
were treated merely as the loci of certain equations, or as the 
graphs of certain functions. We shall next proceed to introduce 
these last curves again as geometric loci; but to facilitate the 
process we shall need to be able to use different sets of coordinate 
axes in handling the same curve. 

50. Transformation of Coordinates. 
Suppose we know the equation 
f \x, y) = of a curve referred to a 
pair of rectangular axes XOY and 
wish to find its equation F(x', y) 

referred to a pair of rectangular 
axes X'O'Y' parallel to these, the 

new origin 0' being the point (x , 

y ) in the old system. Let P be any 
point of the curve, its coordinates 
being (x, y) in the original system, 
and {x f , y') in the new system. 

Then it is evident from Fig. 12 
that the relations between the old 
and the new coordinates are 

5 



Shifting the Origin. — 



y' 



!f° 



a 



•JC--H 



-X 



-X 



Fig. 12. 



50 



The Calculus. 



X — Xq ~\~ X } 

Substituting these values, we shall have 

the desired equation. 

51. Rotation of Axes. — Suppose now that, having the equation 
f(x,y)=0 of a curve referred to a pair of rectangular axes XOY , 
we wish to find the equation F(x', y') = of the same curve when 
the axes of reference have been rotated about the origin through 
the angle 6 to the new position X'OY'. 

Y 




Fig. 13. 



Let P be any point of the curve, and let its coordinates be 
(x, y) in the original system and (af,i/) in the new system. 
Then it is evident from Fig. 13 that 

x—x' cos 6— y' sin 9, 

y—x' sin O+y* cos 6. 

The principal object of transformation of axes is to simplify 
the equation of the curve. It will be found that shifting the 
origin makes no change in the terms of highest degree. 



Analytic Geometry. 51 

As an example, suppose that, having the equation 
x 2 -xy + y 2 + 2x-3y + 2 = 0, 

we use a new pair of axes parallel to the old axes, having the 
point (—-J, t) of the old system (the center of the ellipse) as 
new origin. The new equation, when the primes are dropped, is 

x 2 -xy + y 2 = i. 

If we now turn the axes about the new origin through the 
angle 45°, the equation (after dropping primes) is 

x 2 + 3y 2 = l 

52. Examples. 

1. Shift the origin of the equation 

Sx 2 - ±xy + by 2 - 8x - 16y - 16 = 

to the point (1, 2), and transform the equation thus derived by 
rotating the axes through the angle tan -1 2. 

Ans. ±x 2 + 9y 2 = 36. 

2. Shift the origin of the equation 

6x 2 -5xy-6y 2 -19y-22x+5 = 

to the point (1, —2), and rotate the axes of the equation thus 
derived through the angle tan -1 5. Ans. x 2 — y 2 = 2. 

3. Transfer the origin of the equation 

llx 2 -4xy + 8y 2 -50x-52y + ll = 0. 

to the point (3, 4), and then rotate the axes of the derived 

/ 1 \ x 2 y 2 

equation through the angle tan _1 f — =-). Ans. TX+ 9T = 1- 

4. Rotate the axes of the equation x 2 — y 2 — a 2 through an angle 
of 45°. Ans. 2xy=-a 2 . 

53. Conic Sections. — A conic section is the locus of a point 
which moves so that its distance from a fixed point, the focus, 
has a constant ratio to its distance from a fixed line, the 
directrix. 

The constant ratio, which is designated by e, is called the 



52 



The Calculus. 



eccentricity and determines, by its value, whether the curve is a 
parabola, an ellipse, or a hyperbola. 

Half the distance from the focus to the directrix is called the 
parameter, and is designated by p. The parameter determines 
the size of the conic, the eccentricity its shape. 



*£L 



<#/ 



"A 

/ ! 



F 



M 



Fig. 14. 



If we take as rectangular axes the directrix and a straight line 
through the focus, as in Fig. 14, then by definition, FF—ex. As 
2p is taken for the fixed distance OF we have, from the right 
triangle FPM, 

(x-2p) 2 + y 2 = e 2 x 2 , (1) 

the general equation of the conic. 

When equation (1) is rearranged in the form 

ax 2 + bxy + cy 2 + dx+ ey +f=0, 
as 

(l-e 2 )x 2 + y 2 -±px+4:p 2 = 0, 

the discriminant is seen to be 

& 2 -4ac = 4(e 2 -l). 



Analytic Geometry. 53 



Hence the curve is: 



An ellipse when e<l. 
A parabola when e — \. 
A hyperbola when e>l. 

54. Properties of the Conies. — The line through the focus 
perpendicular to the directrix is called the transverse axis or 
principal diameter of the conic, and the points where it inter- 
sects the conic are called the vertices. The double ordinate 
through the focus is called the latus rectum. 

To locate the vertices we solve equation (1) with the equation, 
y=0, of the principal diameter, getting 

(x-2p) 2 = e 2 x 2 , 
x — 2p= ±ex, 

\+e 1—e 

If e<l, both roots are positive; if e = l, one root is p, the 
other is infinite; and if e>l, one root is positive, the other neg&- 
tive. Hence : 

Both vertices of an ellipse are on the same side of the directrix 
as the focus; 

The parabola has only one (finite) vertex, midway between 
the focus and the directrix; 

The hyperbola has a vertex on each side of the directrix. 

The latus rectum of the general conic (x — 2p) 2 + y 2 = e 2 x 2 is 
^pe, since the ordinate corresponding to the abscissa 2p is ± 2pe. 
In the parabola, this length is 4p, twice the distance from the 
focus to the directrix; in the ellipse, it is less; in the hyperbola, 
greater. 

55. Typical Equations. The Parabola. — When e = l, the gen- 
eral equation (1) becomes 

tf = ±p(x- V ). (2) 



54 



The Calculus. 



To simplify equation (2), shift the origin to the vertex (p, o), 
as shown in Fig. 15. The equation then becomes 

f-^x, (3) 

the typical form of the equation of a parabola, the coordinate 
axes being the principal diameter and the tangent at the vertex. 




Fig. 15. 



56. 



Examples. 



1. What is the equation in typical form of the parabola which 
has the double ordinate 2b corresponding to the abscissa a? 

11 x 

Ans. -tt = — . 
o 2 a 

2. Write the equation of the parabola, vertex at origin, when 
the directrix is x=p; when it is y—p; when it is y— —p. 

Ans. y 2 —— £px; x 2 =— 4=py; x 2 — ^py. 

3. Show that the line y — mx always cuts the parabola y 2 — ^px 
in two points. What are the points when m — ? 

4. Find the equations of the following parabolas: 
(a) Vertex (1, 2) focus (-2, 2). 

Ans. (v-2) 2 = -12^-1). 



Analytic Geometry. 55 

(6) Vertex (2, 3), focus (4, 3). 

Ans. tj 2 -6ij-8x + 2o = 0. 

(c) Vertex (2, 2), focus (2, -4). 

Ans. x 2 — 4x + 24?/ — 44 = 0. 

[Hint: Write the equation of each referred to its transverse 
axis and tangent at vertex; then shift to the required axes.] 

57. The Central Conies: Ellipse and Hyperbola. — When e is 
not=l, the general conic 

(x-2p) 2 + y 2 = e 2 x 2 (1) 

has, as we have seen, two vertices, at 



(&••)-(&•)• 



The line joining the vertices is called the major axis of the 
conic; its length is denoted by 2a. The point midway between 
the vertices will be seen later to bisect every chord through that 
point; it is therefore called the center. (See Figs. 16 and 17.) 

58. The Ellipse. — The length of the semi- major axis for the 
ellipse is 

«_*( _3E-__2£-}_ _3l* , 

2 \ i_ e 1+e/ 1 — e 2 ' 

and the distance from the focus- to the directrix, in terms of this 
length, is 

ail — e 2 ) a 
c e e 



The abscissa of the center is 

x ( 2p 2p\_ 2% 

(See Fig. 16.) 



2p \ 2p a_ 

e 



56 




Equation (1) is much simplified when the origin is shifted to 
e cen 
getting 



the center. If we first replace 2p in (1) by its value— — a e, 

6 



(it-l+fl^ + ^r^ 



and then shift the origin to ( — , Oj, we have, after dropping 
(x + ae) 2 + y 2 = e 2 (x + — Y, 



primes, 



or 



or 



(l-e 2 )x 2 + y 2 = a 2 , 






tr 



a 2 ' a 2 (l-e 2 ) 



(4) 



Analytic Geometry. 



57 



59. The Hyperbola. — The length of the semi- major axis of 

the hyperbola is 






and the distance from the focus to the directrix, in terms of this 
length, is 

a(e 2 — 1) a 
2p= — * ^ =ae . 

The abscissa of the center is _f 2 = — . (Fig. 17.) Eeplac- 

i. e e 




Fig. 17. 



ing 2p by ae— — , shifting the origin to ( — — , J , and drop 
ping primes, we have again 



a? + a 2 (l-e 2 ) 



= 1. 



(4) 



58 The Calculus. 

This one equation thus represents either an ellipse or a hyper- 
bola according as e<l or e>l. 

60. Symmetry of the Central Conies. — It is now evident from 
equation (4) why the conies for which e is not=l are called 
central conies, and why the point taken as our new origin is the 
center. For, corresponding to any point (x 1} y x ) lying on (4) 
is another point (— x lf — y x ), also lying on (4), and as the new 
origin is half-way between the two points, it bisects any chord 
drawn through it. Moreover, the points (x 19 — y ± ) and (—x x , yL) 
also lie on (4), so that the new axes of coordinates are axes of 
symmetry for the conic. These axes of symmetry are called the 
principal axes or principal diameters of the central conic: the 
one perpendicular to the directrix being the major axis, and the 
one parallel to the directrix, the minor axis, or conjugate axis. 

61. Typical Equations of Ellipse and Hyperbola. — The inter- 
cepts on the minor axis, x = 0, of a central conic 



2 y 2 

r _2 / 1 _2\ - 1 - 



el j_ y 

a 2 ^ a 2 (l-e 2 ) 



are ±aVl — e 2 , real in the case of the ellipse, imaginary for the 
hyperbola. 

In the ellipse, the length of the semi- minor axis is denoted 

by b; 

b = a\/l-e 2 (e<l), 

so that we have as the typical equation of the ellipse referred to 
its principal diameters as coordinate axes: 

■$■+-£-!■ < 5 > 

In the hyperbola, a 2 (l — e 2 ) is represented by — o 2 , 






b = aVe 2 -l (e>l), 

so that we have as the typical equation of the hyperbola referred 
to its principal diameters as coordinate axes : 



Analytic Geometry. 59 

(6) 









x 2 y 2 
a 2 b 2 " 


I. 




In this f orm 


the 


asymptotes of the 


hyperbola are 






X 

a 


--f = Oand — 
o a 


4-JL_o 


01 






b 

J a 





In the hyperbola, then, b is the length of an ordinate of either 
asymptote drawn from either vertex. For convenience in certain 
statements, b is often called the semi- minor axis of the hyper- 
bola. 

62. Dimensions of the Central Conies. — From what has al- 
ready been said, the values of the following are evident : 

For any central conic having a major axis of length 2a and 
eccentricity e, we have the lengths : 



Center to directrix : 



e<l. e>l. 

Ellipse. Hyperbola. 

a a 

e e 



Focus to directrix : ae. ae . 

e e 

Center to focus: ae. ae. 

Latus rectum: 2a(l — e 2 ). 2a(e 2 — l). 

Minor axis: b — a\/\-e 2 . [b = aVe 2 -l]. 

In terms of a and b, the eccentricity is : 



e= J a ^=^- for the ellipse, e= J°^- for the hyperbola. 

On account of the symmetry of the curves, the ellipse has a 

directrix — to the right of the center, and a focus ae to the 

e ° 

right in addition to the directrix and focus at the same distances 
to the left ; the hyperbola has a pair to the left as well as one to 
the right. 



60 The Calculus. 

63. Examples. 

1. The focus of a conic is £ in. from its directrix, and the 
eccentricity is f . Find the equation referred to the directrix and 
a perpendicular through the focus, and reduce to the typical 
form. Draw a figure to scale, showing the foci, directrices, cen- 
ter, latus rectum, and minor axis. 

2. Treat the same problem as that of example 1, changing the 
eccentricity to f and replacing the minor axis by the asymptotes. 

3. Determine the dimensions of the following curves : 

(1) 9z 2 + 25^ = 225. Ans. e = i, etc. 

(2) 9x 2 -16f = U4. Ans. e=-J, etc. 

(3) 9x 2 + 25i/ 2 + l&z- lOOy- 116 = 0. Ans. e = $, etc. 

(4) \f— (a — x) (x + a). 

(5) Write the equations of the circles, radius a, when the 
center is: 

(a) at (-a, 0), (b) at (a, 0), (c) at (0, -a), (d) at (0, a). 

4. Prove that a circle having its center at an extremity of the 
minor axis of an ellipse and its radius equal to the semi- major 
axis will intersect the major axis at the foci. 

5. Prove that a circle concentric with a hyperbola and passing 
through the intersections of the asymptotes with the tangents at 
the vertices will also pass through the foci. 

6. Show that the circle is a special case of the ellipse, having 
equal semi-axes, the foci coincident with the center, eccentricity 
zero, parameter infinite, and directrix at infinity. 

7. Suppose a right circular cone cut by a horizontal plane in 
a circular section. If the plane turns about a diameter of the 
circle, how will the eccentricity of the section vary ? What is the 
eccentricity of a pair of intersecting straight lines regarded as a 
conic ? 

64. Conjugate Diameters. — The locus of the middle points of 
a set of parallel chords in a conic section is a diameter. This is 
proved as follows: Let the equation of the conic be 

(x-2p) 2 + y 2 = e 2 x 2 , (1) 

as in Art. 53, and let the common slope of the parallel chords be 
m, and the general equation of all the chords be 

y=mx-rC, (2) 



Analytic Geometry. 61 

m thus being a fixed constant and c a quantity that is constant 
for any one position of the chord (2), but varies as the chord 
moves. (A quantity of this sort is called a parameter.) 

The extremities (x lf y x ) and (x 2 , y 2 ) of any one of the chords 
are the two intersections of (1) and (2), so that x x and x 2 are 
the two roots of 

(x-2p) 2 + (mx + c) 2 = e 2 x 2 , 

or of 

x 2 (l-e 2 + m 2 ) + 2{mc-2p)x + (4p 2 + c 2 ) =0. (3) 

If ¥(x' , y') is the mid-point of this chord, 

t X-, -f- Xo 

and is half the sum of the roots of (3) . By the Theory of Quad- 
ratics (Algebra, Art. 11), this is 



,_ 2p — mc 
X -l-e 2 + m 2 ' 



so that 



c=^-(e 2 -l-m 2 ) + ^.. 
m m 

Since P' lies on the line y=mx + c, we have 

y' — mx' + c, 

or 

y f = mx '+ — (e 2 -l-m 2 ) + ■?£-. 

This relation between the fixed constants and the coordinates 
of P 1 , since it is free from the parameter c, expresses the restric- 
tion under which the point P f moves, or is the equation of the 
locus of P'. Simplifying the equation and dropping primes, we 
have 

y=-^-[(e a -l) ! r+3p] ) (4) 

as the equation of the locus of the mid-point of a set of chords 
of slope m in the general couic. 



62 The Calculus. 

If the conic is a parabola, e = 1, and the locus is 

" = -: . (5) 

a line parallel to the transverse axis. Any such line is called a 
diameter ; if we regard the parabola as a conic having one vertex 
at infinity, and therefore its center also at infinity, we may say 
that all these parallels pass through its center. 

If the conic is an ellipse, its center is at f — , J and 

2p=^-(l-e 2 ); 
equation (4) may therefore be written 



H-t) 



and, referred to central axes, becomes 

p 2 — 1 
y=- — -*■ (6) 

If the conic is a hyperbola, its center is at f — — , Oj and 

2p= — (e 2 — 1) ; equation (4) may be written 
c 

e 2 -l( , a\ 

and, referred to central axes, becomes 

y- x. (6) 

For any central conic, then, the locus of the middle points of a 

e 2 — l 
set of parallel chords, of slope m, is a line of slope through 

the center, or a diameter. 

The locus of the mid-points of chords parallel to the diameter 
(6) is y—mx, one of the original set of parallel chords. This 

e 2 — 1 
is seen by putting ■ for m in the italicised result of the pre- 
ceding paragraph. 



Analytic Geometry. 63 

Thus we have the theorem: If the product of the slopes of 
two diameters of a central conic is (e 2 — 1), each of the diameters 
bisects all the chords parallel to the other. 

Two such diameters are called conjugate diameters. 

65. Focal Radii of Central Conies. — The straight lines joining 
the point P(x, y) of a central conic, as in Figs. 16 and 17, with 
the foci F and F 19 are called focal radii. In the case of the 
ellipse, these distances are by definition : 

FP=e(—+x\=a + ex J 

and 

F 1 P=e(—-x\=a-ex; 

their sum is therefore constant, and equal to 2a. 
In the case of the hyperbola, the focal radii are 



and 



FP—e (x — —J —ex — a 
F 1 P=e(x+ -^\=ex + a; 



their difference is therefore constant, and equal to 2a. 

Consequently the ellipse may be defined as the locus of a point 
moving so that the sum of its distances from two fixed points is 
constant; and the hyperbola as the locus of a point moving so 
that the difference between its distances from two fixed points 
is constant. 

66. Examples. 

1. The principal axes of a central conic divide the plane into 
four quadrants. Show that two conjugate diameters of an ellipse 
never lie in the same quadrant, but two conjugate diameters of a 
hyperbola always do. 

2. What is the equation of the diameter conjugate to y=2x 

x? xp x^ ip 
in the conic -q- + -4- = 1 ? In the conic q -4- = 1 ? 

Ans. y=-ix; y= - g - . 



64 The Calculus. 

3. Show that if a diameter y — mx meets a hyperbola 

s 2 J/1 _ 1 
a 2 b 2 ~ l 
in real points, the conjugate diameter does not. 

4. Two pins are stuck in a piece of paper and an endless string 
is held taut about them by a pencil point. What curve will be 
drawn when the pencil moves ? Find its dimensions if the string 
is 7 in. long and the pins are 3 in. apart. 

Ans. a=2, e = f, &=-i\/7, etc. 

5. Show that if the extremities of a diameter of a central 

x 2 ?/ 2 

conic -^2 + a2 ^_ e2 , = lave (x^y^ and (-x ± , -y ± ), the extremi- 
ties of the conjugate diameter are 

(yfep -^VW) and (^=, ^VI=?). 

67. Tangents, Normals, Subtangents, and Subnormals. — We 

have already seen that if a curve is given in rectangular co- 
ordinates by an equation f(x, y) =0, the slope of its tangent at 

any point is the value of -?■ at that point. If P (Fig. 18) is any. 

point of the curve, draw any horizontal length from P to repre- 
sent dx, and a vertical length in proper proportion to represent 
dy, and complete the triangle by drawing the line marked ds. 
The resulting triangle is called the differential triangle for the 
point P. 

Let PT and PN be the tangent and normal to the curve at P, 
meeting the axis of x at T and N respectively, and let PM be the 
ordinate of P. Eepresent the angle PTM as usual by r, and note 
that the angle MPN and the angle from dx to ds are each equal 
to t, as all the right triangles in the figure are similar. 

We are to find the equations of the tangent PT and the normal 
PN, and the lengths PT, PN, TM, NM. These four lengths are 
called : PT, the tangent; PN, the normal; TM, the subtangent; 
and NM, the subnormal of the curve for the point P. Abbreviate 






Analytic Geometry. 



65 



the four required lengths as shown in the figure, and let the 

coordinates of P be (a, b). Determine the general expression 

dv 
for -j- for the curve; then 



dy 



dx 



= tan t= 



J x—a 



St 



sn 
T 



Y 




'? 


7Vt\ o 


< I a 


A 


\ x 


St. 


„ * 


j, an. 


7f 



Fig. 18. 



Hence 



L*J. S UJ™' 

Since cfc= V(<fcr) 2 + (<fy)*, 



68 The Calculus. 

The equation of the tangent is y — &=tanr(a;— a) 
or 



-4-t], 



(x-a\ 



a 



The equation of the normal is y — b= — cot r (x—a) 
or 

'-'Hi!..-"-"- 

This discussion exhibits the general methods of finding the 
two equations and the four lengths. In any numerical case, the 
simplest process is to sketch a figure similar to Fig. 18, mark the 
values of a, b, &x, dy, and ds for the given point, and derive the 
results directly from the similar triangles. 

68. Application of Art. 67 to the Conies. Parabola. — Given 
any parabola, of parameter p, refer it to its transverse axis and 
the tangent at its vertex as coordinate axes. Its equation is then 

y 2 = ±px, 

and its slope at any point is 

dy __ 2p 
dx ~ y 

If P ± (x ± , y x ) is any point on the parabola, the tangent at 
that point is 

or 

y 1 y-2px=y 1 2 -2px 1 = 2px 1 , 

since, the point P ± being on the parabola, y 1 2 = 4:px 1 . 

The tangent to y 2 = 4px at any point (x 1} y x ) lying on the 
curve is y 1 y=2p(x+x 1 ). 



Analytic Geometry. 67 

The normal at P 1 is 

or replacing x ± by its value in terms of y 19 

y 1 x + 2py=^(8p 2 + y 1 2 ). 

For the differential triangle, if we take dx=y 19 we have 
dy-2p, ds=YIf + yJ. 

The subtangent = y 1 -^- = -|^ = ^ =2^. 

The subnormal — y, — " =9p. 

The tangent = y x ^ 4p " + yi2 =2Vx 1 (x 1 +p). 

6p 



-„ V±P 2 +yi 



The normal — y x v r tvi — 2^p(x ± + p). 

Note that x x is the distance of P x from the tangent at the 
vertex, p is the distance of the vertex from either the directrix or 
the focus., (x ± + p) is the distance of P x from the directrix. 

The Central Conies. — The typical equation of any central conic 
(ellipse or hyperbola), in which the coordinates are measured 
from the principal axes, is 



From this equation, 



^ = (« 2 -D- 

ax v ' y 



In the differential triangle at any point P ± of the conic, if we 
take dx=y 19 then 

dy=(e 2 -l)x 19 
and 

ds=Vy 1 2 +(e 2 -l) 2 x 1 2 =\/(l-e 2 )(a 2 -e 2 x 1 2 ). 



68 The Calculus. 

Prom these, remembering that P x is on the conic, we derive the 
following : 

Equation of tangent : -K + 0/ x ox = 1. 
u ° a a 2 (l — e 2 ) 

Equation of normal: 1- (e 2 — 1) -*- = e 2 . 

x i Vi 

Subtangent = — . 

x i 

Subnormal— (e 2 — l)x 1 . 



1 



Tangent = — V ( a 2 — x 2 ) ( a 2 — e 2 x ± 2 ) . 
x i 



Normal=V(l-e 2 )(a 2 -e 2 x 1 2 ). 

No attention is paid to the signs of the last four, as merely 
the lengths are of interest. The absolute value of each of them 
is to be used. 

The quantity (a 2 — e 2 x 2 ) is the product of the focal radii of 

Pv 

For the typical equation of the ellipse and hyperbola, a 2 (l — e 2 ) 

is replaced by b 2 and — b 2 respectively; so: 

The tangent to the ellipse -y + ^- = 1 at any point (x 1} y x ) 

lying on the ellipse is -—- -f- -^ = 1. 

x^ %P 
The tangent to the hyperbola — r — -fj = 1 at any point (x ly y x ) 

Cb 

lying on the hyperbola is -\ — ^M-=l. 

It can be shown that the equation of the tangent to any conic 
at any point (x 19 y x ) lying on the conic can be written by first 
writing the equation of the conic with xx in place of x 2 , yy for 

y 2 > i( x y+y x ) f° r x v> i( x + x ) f° r x > an ^ i(y+y) f° r y> an( i 

then affixing the subscript 1 to the alternate letters. 



Analytic Geometry. 69 

69. Tangents and Normals to the Parabola in Terms of the 
Slope. — The tangent to y 2 — \px may be written: 

2p „ , 2px, 
or, since y 2 — ^px x , 

U 2/i 2 

If we represent the slope of the tangent by m, 

2p 
Vi 

2 ~ m ' 

so that the equation of the tangent of slope m to the parabola 
y 2 = ±px is 

J m 

The normal to the parabola y 2 — 4,px at the point P x of the 
parabola may be written : 

If we represent the slope of the normal by m, 
y 1= -2pm; 

so that the equation of the normal of slope m to the parabola 
y 2 = 4:px is 

y = mx—2pm — pm s . 

Note the significance of these forms ; if we draw a line of any 
slope m and make the ^-intercept = p/m, the line will be tangent 



70 The Calculus. 

to the parabola y 2 — ^px; if the ^-intercept is made = —2pm 

— pm 3 , the line will be perpendicular to the parabola. 

These forms furnish a means of determining the tangents or 
normals to the parabola from any point, whether it lies on the 
parabola or not. 

For instance, to find the tangent from (4, 7) to y 2 = 6x, since 
any tangent is in general 

and the desired tangents pass through (4, 7), we have 

solving this for m we get m—\ or \. Hence the tangents are 
2y-Zx-2 and 4y-a;=24. 

The slopes of the tangents from (x 1? y x ) to y 2 = 4=px are in the 
same way i(y 1 ± Vy^ — ^px x ) ; both are real for a point outside 
the parabola, they coincide for a point on the parabola, and are 
imaginary for a point inside the parabola. 

To find the normals from (12, 6) to y 2 = 6x; since they are of 
the form 

y = mx — 2pm — pm? 

and pass through (12, 6), we have 

6 = 12ra-3m-fra 3 

to determine m; that is, the roots of m 3 — 6m+ 4=0 are the 
slopes of the required normals. These roots are 2, V3 — 1, and 

- V3-1 or 2, .732 and -2.732. 

There are thus three normals from this one point to the para- 
bola. The equation for the slope, m, of a normal from (x lf y t ) 
to y 2 = 4px is similarly seen to be 

m 3 + 2p ~ Xl m+ -&- =o. 
P V 



Analytic Geometry. 



71 



According as the discriminant of 
this equation (Algebra, Art. 74) is 
negative, zero, or positive, there will 
be three different normals, two co- 
incident and one different, or one 
real normal and two imaginary. 

[The discriminant is 

1 ±^-(27py^ + 4:[2p-x i r), 

so that the curve 27py 2 = 4c(x — 2p) 5 , 
called a semi-cubical parabola, is the 
boundary of the points from each of 
which three real normals can be drawn to the parabola.] 

70. Tangents to the Central Conies in Terms of the Slope. — 




The tangent to - 3 + -^ 



1 1S a 2 + 6 3 



ViV - 



1 or 



b 2 x x b 2 

y— — g-^H — 



If we represent its slope by m, 



and since 



whence 






m= ± 



t (6 f -yi f ); 



<*Vi 



Vb 2 - Vl *; 



— = ±Va 2 m 2 -f b 2 : 

so that the equation of the tangent of slope m to the ellipse 

t 

b 2 



2 n? 



y=mx± Va 2 m 2 + b 2 . 



72 The Calculus. 

In the same way, we find that the equation of the tangent of 

x 2 ii 2 
slope m to the hyperbola —j — —- — 1 is 



y = mx ± V a 2 m 2 — b 2 . 

For the circle x 2 + y 2 = a 2 , a special case of the ellipse, the 
similar equation is 



y=mx±aVl + m 2 

(Algebra, Art. 117). 

These equations are of the same service as those for the para- 
bola ; the equations of the normals of the central conies in terms 
of m are less interesting, and much more complicated. 

71. Properties of Tangents to Conies. — The focal radius and 
diameter through a point of a parabola make equal angles with 
the tangent at that point. 

The focal radii to a point on a central conic make equal angles 
with the tangent at that point. 

It is in consequence of the first of these properties that a beam 
of light parallel to the principal axis of a parabolic mirror con- 
verges at the focus, and that rays diverging from the focus are 
reflected as a beam of parallel rays. 

From the second property it follows that rays diverging from 
one focus of an elliptical mirror are brought to a focus at the 
other, and rays diverging from the focus of a hyperbolic mirror 
diverge after reflection as if they had come from the other focus. 

The second property is proved as follows: 

The slope of the tangent to the conic 

a* ' a?{l-e'j~ ' 

is 



X 2 


+ 


f - 


"a 2 


a 2 (l-e 2 Y 


m x — 


:(e 2 -l)^ 
2/i 



Analytic Geometry. 73 

The slope of the focal radius joining (x x , y x ) to (ae, 0) is 



m,- ** 



x x — ae 
The tangent of the angle between these lines is 

m 2 — m 1 _ x x — ae y x Vi 2 + (l — e 2 )x x (x x — ae) 

l+m 2 m 1 " 1 _/ 1 _ e 2\^i . Vi ~ yi(si-ae) ~ (l-e 8 )giyi* 

Since yi 2 = (1 — e 2 ) (a 2 — #i 2 ), this reduces to 

flCl-e 2 )^-^) = fl(l-e 2 ) 
-ey^a-exj ey x 

The slope of the other focal radius is 

m-o = — ^— , 
a^ + ae' 

so that by changing e to — e throughout the preceding discussion 
and reversing the order of the lines, we find 

m x — m s _ _ q(l — e 2 ) 

l + m x m 3 ~ ey x 

The angles are thus shown to be equal. 

72. Perpendicular tangents of a conic intersect in a circle 
called the director circle, which in the parabola is a straight 
line, the directrix (a circle of infinite radius). This may be 
shown in two ways. For example, if the equation of the tangent 
to the ellipse 

y=mx± Va 2 m 2 + b 2 
be arranged as a quadratic equation in m, it will give 

m 2 + 2m -^L+^^J =0. 
a 2 — x 2 a 2 — x 2 

In order that the two direction ratios m x and m 2 derived from 

this equation shall represent tangents at right angles to each 

other, they must have the relation 



74 The Calculus. 

m x m 2 — — 1; 

that is, the absolute term of the quadratic 

b 2 -y 2 i 

a 2 — x 2 * 2 ' 

which gives 

a; 2 + 2/ 2 = a 2 + & 2 . 

For the parabola, we should have, in a similar way, 

m 2 — m — + — = 0, 
x x 



or 

P 
or 



m m 2 = ^— = — 1. 

1 2 a; 



x=—p. 

The same results are readily obtained by finding the locus of 
the intersection of two perpendicular tangents : 



mx — V a 2 m 2 + b 2 , 



my + x= Va 2 + b 2 m 2 . 

If the two equations of these perpendicular tangents are squared 
and added, there results 

(x 2 + y 2 ) (1 + m 2 ) = (a 2 + b 2 ) (1 + m 2 ) ; 

and by eliminating m, we obtain 

x * + y 2 = (i 2 + l 2 . 

73. Pedal Curves. — Perpendiculars from the focus to the tan- 
gents of a conic meet on a circle. 

The equations of the tangent of the ellipse and the perpen- 
dicular from the focus upon it are 



y — mx— V a 2 ra 2 + b 2 , 
my + x = ae. 



Analytic Geometry. 75 

Squaring and adding these equations and noting that o 2 — 
a 2 (l — e 2 ), we get 

(x 2 + y 2 )(l + m 2 )=a 2 (l + m 2 ), 

or, eliminating m, 

x 2 + y 2 = a 2 , 

as the locus of the intersection of the two lines. 

This circle is called the major auxiliary circle of the ellipse, 
for reasons that will appear in Art. 87, Parametric Equations. 

In the case of the parabola, m may be directly eliminated; the 
equations are: 

u—mx-\- -P- , 
v m 



Subtracting, 



or 



x v 
* m ^ m 



x { m +i)' 



x=0. 



The interpretation of the tangent at the vertex of the parabola 
as corresponding to the auxiliary circle of the ellipse is evident 
from the fact that the center of the parabola is infinitely distant 
from the vertex. 

The locus of the intersection of a tangent of a given curve 
with the perpendicular from a given point is called a pedal 
curve. We have just derived the pedals of the conies when the 
given point is the focus. The pedal for any other point is derived 
in a similar way; the equations of the tangent and of the per- 
pendicular from the given point are written in terms of m, and 
the arbitrary parameter m eliminated. 

As an illustration we will derive the pedal curve of the para- 



76 The Calculus. 

bola \f — \ax, when the given point is (— a, 0). The equations 
of the tangent and of its perpendicular from (— a, 0) are: 

V= mx +-^> (1) 

y=^ ±a - (2) 

The elimination of m and the resulting equation are both 
simplified by shifting the origin of (1) and (2) to the given 
point ( — a, 0). The lines are then 

y = m(x-a) + -^, (1)' 

y=-— • (2)' 

/>• 

Putting for m in (1)' and simplifying, we get 

or(a; 2 + 2/ 2 )-a(^-2/ 2 )=0 

as the equation of the pedal curve referred to the given point as 
origin. 

74. Examples. 

1. Find in detail the equations of the tangent and normal and 
the lengths of the tangent, normal, subtangent, and subnormal 

X 1] 

for the ellipse —^ + -—■ — 1 at the point (x 1} y x ) lying on the 

Oi 

curve. 

2. Give the equations and lengths of the tangent and normal 
and the lengths of the subtangent and subnormal to each of the 
following at the point indicated. 

y 2 = 12x at the point having the abscissa 3. 

— 4- -~ = 1 at the point having the abscissa 2. 

— — -^- = 1 at the point having the abscissa 3. 



Analytic Geometry. 77 

3. Find the tangents from (2, 7) and the normals from 
(15, 6) to the parabola y 2 = 12x. 

Ans. y = 3x+l, 2y = x+12, y = x — 9, y— -2x+36. 

or ifi 

4. Find the slopes of the two tangents to — -j- -^- = 1 from 

(A, 2) ; from (3, Je) ; from (3, 2). (See Algebra, Art. 46.) 
Ans. y^- and 0; ^^ and oo ; and oo. 

5. Find the equations of tangent and normal and lengths of 
subtangent and subnormal for the curve y — e x where it crosses 
the y-axis. Ans. y — x=l, y + x=l, 1, 1. 

6. Find the lengths of tangent, normal, subtangent, and sub- 
normal for y — e x , where x = log V3. Ans. 2, 2V3, 1, 3. 

7. Show that the lengths of subtangent and subnormal for the 
curve y — sin x are in general tan x and \ sin 2x. 

8. Find the subtangent of y = tan x when x=\ . Ans. J. 

9. Find the angle between y — sin x and ?/ = coS£. 

Ans. tan-^v^- 

10. Show that the tangents at the extremities of the latus 
rectum of a conic meet on the directrix. 

11. Let PM be the ordinate, PN the normal of a point P on 
the conic y 2 = (1 — e 2 ) (a 2 — x 2 ), and call the foci F r and F*. 
Show that NF ± : NF 2 = PF X : PF 2 , so that by plane geometry PN 
bisects the angle F 1 PF 2 . 

12. Prove the property of the parabolic mirror. 

13. Derive the pedal of the circle x 2 -\-y 2 — a 2 with respect to 
the point (a, 0). 

Ans. (a, 0) being taken as a new origin, the pedal is the 
cardio id ( x 2 + y 2 + ax) 2 = a 2 (x 2 + y 2 ) . 

14. Derive the pedal of the rectangular hyperbola x 2 — y 2 — a 2 , 
with respect to its center. 

Ans. The lemniscate (x 2 + y 2 ) 2 = a 2 (x 2 — y 2 ), the origin be- 
ing the same as for the hyperbola. 

75. The Second Derivative. — If y is a function of x, the de- 
rivative of y with respect to x is also a function of x, the deriva- 
tive of which may be taken with respect to x. 

The x-derivative of the x-derivative of a function is called the 
second x-derivative of the function, or the second derivative of 
the function with respect to x. 



78 The Calculus. 

The derivative of the second derivative is called the third de- 
rivative, and so on. 

The notation used in expressing the higher derivatives is as 
follows : 

If y=f(x), then 

y' — f(x) = 'y' — -^- represents the first derivative; 
y"— f"{x) = ' y' = —^— represents the second derivative; 

yf" = f"{x) = ' ,\ ' = -jj— represents the third derivative; 

and so on. 

When it is convenient to consider that the value of dx, which 
is arbitrary, is always the same throughout a discussion, another 
set of forms can be used to denote the higher derivatives. For if 

dx is considered constant, and we find the x-derivative of ~- by 
differentiating and dividing by dx, we get y" — }\ -^ ; repeating 

the process, we get y'" — / ) ffl ? and so on. 

These numerators are cumbersome, so they are abbreviated to 
d 2 y, d 3 y, expressions which are read " d second y " or " second 
differential of y" etc., and which mean " the differential of the 
differential of y," etc. 

The notation, when x is the independent variable, is thus : 

y=f(x), 

•= r («>=-£' 



tf'=f"{*)=^. 



etc. 



Analytic Geometry. 79 

It will be noticed that in place of (dx) 2 we have written dx 2 , 
which in these forms is always taken to mean " the square of dx." 
Whenever there is any danger of confusion with the differential 
of x 2 , the parentheses must be retained. 

The second derivative is the only one that is susceptible of 
interpretation by itself, though the others occur very commonly. 
The simplest interpretation of the second derivative occurs in 
connection with functions of time. If s = f(t) is the distance 
traversed during the time t, 

s'=f(t) = f, 

is, as we already know, the speed of the motion, or the time-rate 

d 2 s 
of increase of the distance, s" = f" ( t) — -rry- is called the ac- 
celeration of the motion, and is the time-rate of increase of the 
speed. Thus when a body falls to the earth, going s ft. in t sees, 
the law of its falling is (nearly) 

s = lQt 2 ; 
from which 

or the speed is 32i ft. a second; and 

*"=-§- =33 w, 

or the acceleration is 32 ft. a second in each second; that is, there 
is an increase in the velocity of 32 ft. a second during each second 
that the body falls. 

In problems of motion, then, the first derivative is the speed, 
the second derivative is the acceleration. The third derivative 
occurs in some sorts of motion, but has no special name. 

In geometrical problems in which a curve is represented by an 
equation y=f(x) in connection with a system of axes, the first 



80 



The Calculus. 



derivative, -^ , is the slope of the curve, as we have seen. The 

second derivative has no special name, but is very significant, both 
by itself and as an element of an important expression. 

76. Inflections. — Let y — f(x) be the equation of a curve in 
rectangular coordinates; then 

2/ '=/'(*) = |f =tanr 
is the slope of the curve, and 

is the derivative of the slope, and is consequently positive when 
the slope is increasing (algebraically), negative when the slope is 
decreasing, and zero when the slope is changing from one state to 
the other. A point which appears to be the junction of opposite 
bends, like the middle point of the letter S, evidently has this 




Fig. 19. 



Analytic Geometry. 81 

last property; such a point is called a point of inflection, and at a 
point of inflection of a curve, y = f(x), 

jr=r<*)=-g-=o. 

Any graph of a cubic function, such as we studied in the 
Algebra, has just one point of inflection. Thus the equation of 
the curve on page 99 of the Algebra is 

10y = 2x 3 -15x 2 + 24:X+6. 

And for this curve, 

10ij' = f (x) = 6z 2 -30a? + 24, 
10y" = f'(x)=12x-30, 
y" — when a?=|-. 

As x increases from — oo to + oo , the slope y' decreases, as is 
evident from the figure, from + 00, becoming zero when a?=l, 
decreasing further to the value =^f when a?=§ (at the inflec- 
tion), after which it increases to + oo, passing through the value 
zero when x=4=. 

In this figure the inflection is clearly the junction of the part 
of the curve convex upward with the part convex downward, so 
that the tangent at the inflection crosses the curve. Inspection of 
an adjacent secant shows further that at the inflection the curve 
and the tangent have three common points. This appears also 
analytically from the fact that the equation of the tangent, at the 
point (|, £$), 

1n 149 -54a? 

i<>y= — ^ — , 

when solved simultaneously with the equation of the curve, gives 

(2a?-5) 3 = 

to determine the intersections. 
7 



82 The Calculus. 

77. Examples. 

1. Show that the tangent to the inflection for y—x? — 3x* 
-45#+7 is48z+2/=8. 

2. Show that if the abscissas of the highest and lowest points 
of a cubic parabola a 2 y=x 3 + bx 2 + c 2 x+d 3 are x ± and x 2 , the 

abscissa of the inflection is ^^ . 

3. Find the inflection of y=x 3 + px+q. 

4. Find the tangent to the curve y(l+logx) —x at its inflec- 
tion. Ans. ^y—x — e. 

5. Find the tangents at the inflections of the curve (Witch) 

y(a?+x 2 ) = a 3 . 

Ans. 2y±x = 2a, 

6. Show that if y=x~ 3 *> . VL = _ 61o g^ , ^"~/ 2 = 

2/ x ' y- 

--Jr(l-log*), <= — 5-(l+»log.*)(l-31og*),BO that 
the curve y—x~ 3 l09 x has inflections at {e~^, e~%) and (e*, e~^). 

78. Curvature. Radius of Curvature. — One of the most im- 
portant characteristics of a curve, especially in connection with 
the bending of beams and other structural supports, machine 
parts, etc., is the sharpness with which the curve bends, or its 
curvature. 

Some curves bend more sharply than others, and except for 
the straight line and the circle, any curve bends more sharply 
at some points than at others ; the curvature of any conic section, 
for instance, is greatest at the ends of its major axis, and de- 
creases at increasing distances from, these points. 

To measure the curvature of any given curve at a given point, 
P ly let the tangent at P t slide along the curve, its point of con- 
tact moving through the arc P ± P 2 = As, and the tangent itself 
turning through the angle Aa (Fig. 20). The angle of turn, Aa, 
in comparison with the distance, As, that the point of contact 
has to move in order to produce this turn, indicates the sharp- 



Analytic Geometry. 



83 



ness of the curvature at P ± . The quotient, 
- -"— , is the mean curvature of the curve 



As 

from P 1 to P 2 . This mean curvature 
varies, in general, for the same point P 1? 
according to the length of As. As usual 
in such cases, its value when As=0 is taken 
as the actual curvature at P, ; i. e., 




Aa 

As 



Fig. 20. 



As=0 



X, 



the curvature at P ± . 



79. Given a circle of radius a> to find its curvature at a given 
point P 1 (Fig. 21). Choose any second point, P 2 , and call the 
arc P 1 P 2 = As. Call the angle turned 
through by the tangent as its con- 
tact moves from P x to P 2 , Aa. Choose 
any diameter OA and call the angle 
P 1 OA = 6, the angle P 2 OA = 6+A0. 
Then by geometry, 

Aa = A0, 

and using circular measure through- 
out, we have 

As = aA0: 




Fig. 21. 



Aa 

As 



Afl 
aA6 



For the circle, then, the mean curvature is constant for all 
points and for all lengths of the increment of arc ; the curvature 
of a circle at any point is the reciprocal of the radius. 

For any straight line, Aa = for any value of As, and the 
curvature is constantly zero. (This is consistent with the con- 
ception of a straight line as a circle of infinite radius.) 



84 



The Calculus. 



80. To compute the curvature of any curve at a given point, 

let <f> be the angle made by a tangent 
to the curve with any fixed line OA 
(Fig. 22), and A<p be the increment 
<p receives when the contact moves 
through the arc P 1 P 2 = As; then 
Aa — Acf), and the curvature is 




X = 



As ! 



A.s=0 



d<p 
ds 



d<f> . 



The value at P 1 of ~j- is the actual curvature at P t . 

The most convenient method of designating the curvature of 
a given curve is to give the radius of the circle which has the 
same curvature; this radius is denoted by p (rho, the Greek r) ; 

then as the curvature of the circle is — , 

P 

1 _ d<p __ ds 

~y-~aV' p -~d$' 
This circle is called the circle of curvature; its radius p is 
called the radius of curvature, and its center the center of curva- 
ture. 

81. The Differential of Arc. — In order to use this formula in 
connection with curves whose properties are defined by equations, 
we shall need analytical expressions for ds and dcf>. The symbol 
ds is the same that we used earlier for the hypotenuse of the dif- 

ds 

f erential triangle, and is used here because -j— is sec r when ds 



is the differential of arc as well as when ds— V (dx) 2 + (dy) 2 . 
That is, the differential of arc and the hypotenuse of the differen- 
tial triangle are the same. 

From the familiarity that we have already gained with the 
intimate relation between a curve and its tangent, especially with 
their infinitesimal increments, this idea probably will seem an 



Analytic Geometry. 



85 



obvious inference ; if it is at all vague, the following explanation 
should make it clear. m 

82. By the length of a straight line is meant the number of 
times it will contain some recognized standard or unit; as a 
definition for the length of a curved line we have: "the limit 
approached by the length of a broken line formed of chords of 
the curve as the length of each chord approaches zero, and as the 
number of chords consequently increases indefinitely." Xow 
suppose we have two points of a curve, P(x, y) and P'(x + Ax, 
y + Ay), between which is the arc 
As. Let the slopes at P and P' 
be tan t and tan (r + Ar). Sup- 
pose a broken line to be inscribed 
in the arc from P to P', and for 
convenience let the chords of 
which it is composed have equal 
projections along Ax. (See Fig. 
23.) 

The inclination of each chord to the .r-axis is between the 

values r and (r + Ar) ; its length is the length of its projection 

multiplied by the secant of its inclination. If there are n chords, 

Ax 
the length of the broken line is times the sum of all n 

secants, or is Ax times the average of the secants. This average 
is certainly between sec r and sec(T-f-AT) in value; so the length 
of the broken line is between the values Ax sec r and Ax sec 
(r + Ar). This is true independently of the number of the 
chords composing the broken line, and so holds true when this 
number is increased indefinitely; hence the length of the arc As 
is intermediate in value between AzsecT and Ax sec (r + Ar). 
Thus we have : 

As 

The value of—— is between sec t and sec (r + Ar). 

Ax v J 



Y 


«. 




AX >/„. 












\\r 










£ 






~^P 1 




^T ' IT+UT 


(V 










X 



Fig. 23. 



86 The Calculus. 

As Ax approaches zero, sec(T+Ar) approaches seer, and so 

ds 



I &X J 4x=0 



dx =S6CT ' 



where -p is the ^-derivative of the arc s, and ds is the differen- 
tial of arc. 
But 

sec r= VI+giFr= Ji+ (-g-J = V&Y+ W ; 

hence 

ds=V(dx) 2 +(dy) 2 , 
where ds is the differential of arc. 

ds is really ± V (dx) 2 + {dy) 2 ', in any problem in which the 
sign is of importance, it must be chosen consistently with the 
relations 

ds _ ds 

— =— = sec r, — t- = esc r, 
az «?/ 

etc., so that the derivatives shall indicate by their signs whether 

s increases or decreases with x or y and shall be of the same signs 

as the corresponding functions of t. 

83. Curvature in Rectangular Coordinates. — Given a curve 

y=f(x) referred to rectangular coordinates, to find its radius of 

curvature at any point. Take the axis of x as the fixed line OA 

in the preceding article; then 

<£ = T, 

— d<l> _dr _ ds 

X ~ ds - ds 1 P ~ dr ' 

T =tan- 1 -^ ^tan- 1 /- 
Differentiating t, and dividing dr by ds, we have: 

dT - d(Y) 



Analytic Geometry. 87 



Since 



ds=Vdx 2 + dy 2 =Vl + y' 2 • dx, 

dr d(y') 

ds " (l + y' 2 )i-dx' 

dyj' _ » 
dx ~ J ' 

_ dr_ f 

x ~ ds - (1+y' 2 )*' 

_ ds (1 + ^ 2 )' 
P ~ dr ~ y" ' 

If the problem is one in which x is the independent variable 

d 2 v 
throughout, y" may be replaced by -r4 ; the formula is often 

written 

1 + 
?- 



\dx ) 



d 2 y 

dx 2 

We commonly pay no attention to the sign of p; whenever 

there is any point in considering it, the simplest method is to 

ds 
observe that the sign of p = —7— shows whether s and r are in- 



creasing together or not. The value of ds is ± Vl + 2/' 2 * dx 
according as the arc s and the abscissa x increase together or do 
not, and this of course depends partly on the point of the curve 
from which s is measured, and the direction of the curve that is 
considered positive. 

84. Radius of Curvature of the Central Conies. — As an illus- 
tration of the use of the formula, we will find the radius of 
curvature of any central conic (origin at center). 

tf=(l-e*)(a*-x*). (1) 



88 The Calculus. 

Substituting the value of y 2 from (1) in the numerator, we 
have 

2/"=-(l-* 2 ) 2 ^- 

Now 



i W2_ V 2 + (e 2 -l) 2 x 2 _ (1-e 2 ) (a 2 -x 2 ) + (e 2 -l) V 
by the same substitution, or 



f y 2 



l + y' 2 =~ L yf-(a 2 -e 2 x 2 ). 
Hence 

(i±fii_ (1 ? !li(,,! - eV)l 

p ~ y" ~~ 



(!-« 2 ) 2 ^ 



or the positive value of p is 



(a 2 -e 2 x 2 )S = (e 2 ^ 2 -a 2 )l 



For the ellipse, a, V 1 — e 2 = b, and for the hyperbola, aVe 2 — 1 
= &; hence: 

T/ie radius of curvature of the ellipse — ^ + 4- 2 — 1 at any point 
{x, y) is 

T/ie radius of curvature of the hyperbola — j — p- =1 at any 
point (x, y) is 

ir(^-« 2 )-. 



Analytic Geometry. 89 

For either curve, p is thus the f power of the product of the 
focal radii drawn to the given point, divided by the product of 
the semi-axes. 

85. Examples. 

1. Derive the value of p for the parabola y 2 = ±px directly, first 

using ?/'= a/2- , then using y> = -?- . 

x y 

Ans - p=^( x +p) i= ^(y 2+i P 2 y-- 

v x 

2. Find p for the parabola -p" = — . 

3. What is the radius of curvature for a point of inflection? 

-1. Find p for y = cosx J and show that the least value of p is 
for one of the highest or lowest points, when the center of curva- 
ture is on y=0. 

5. Show that at the vertex, p is equal to the semi- latus rectum 
for any conic, and at an extremity of the minor axis of the 

ellipse, p— j- . 

6. Derive the value of p for the ellipse directlv from the equa- 

.. x 2 v 2 . "(a y + &y )* 

kon^ + p = l. Ans. P = J ^ . 

X 1] 

7. Do the same for the hyperbola — 2 - — jT = 1- 

8. Find the normal and the radius of curvature for 

y—\ (e x/c + e- x/c ). 

Ans. p = n=y 2 /c. 

86. Auxiliary Circles of the Ellipse. — Consider the following 
locus problem. Two concentric circles are given (Fig. 24), of 
radii a and b (a>&), and two perpendicular diameters, OX and 
OY; let any radius Op'p cut the smaller circle at p' and the larger 
at p; to find the locus of a point, P, moving so as always to have 
the same abscissa as p and the same ordinate as p', with refer- 
ence to the axes of coordinates XOY. 



90 




Let the angle pOX=<j>; then for any value of <f>, 



■ a cos <f>, 

■ b sin <f>. 



(i) 



Since sin<f> = ^- } coscf>= — , and sin 2 <f> + cos 2 <f> = 1, we have 



a 2 + 6 2 ~ l5 



(2) 



a relation true for any position occupied by P under the given 
conditions. The locus is therefore an ellipse of semi-axes a and 
b. The larger circle, tangent to the ellipse at the extremities of 
its major axis, is called the major auxiliary circle of the ellipse; 
and the smaller circle, tangent to the ellipse at the extremities 
of the minor axis, is called the minor auxiliary circle of the 
ellipse. The points p and p' are spoken of as the points of the 
auxiliary circles corresponding to the point P of the ellipse. 
<f> is called the eccentric angle of the point P. 



Analytic Geometry. 91 

This proposition furnishes a simple means of constructing any 
number of points of the ellipse by the aid of the auxiliary circles, 
and there are other relations through which certain construc- 
tions are made possible. For instance, the subtangent of any 

ellipse having the major axis 2a is — — — . Therefore if any 

number of ellipses are drawn with a common major axis, points 
on them having the same abscissa will have the same subtangent, 
and the major auxiliary circle is one of these ellipses. Thus a 
tangent may be drawn to the ellipse at any point, or from any 
point of OX, by first drawing the tangent to the major auxiliary 
circle for the corresponding point. 

87. Parametric Equations. — Besides these geometric relations, 
and much more important, is the fact that the pair of equations 
(1) represent the same relation between x and y as the single 
equation (2), for in most analytic work it is easier to use equa- 
tions (1) than equation (2). 

Equations (1) express each of the coordinates of the variable 
point as a function of an auxiliary variable, <£, so that by assum- 
ing all possible values of <f>, all possible pairs of values of x and y 
can be found, and the points of the locus determined. An 
auxiliary variable of this sort is called a parameter, and a pair of 
equations such as these are called parametric equations of the 
corresponding curve. 

It is often convenient to replace the ordinary equation of a 
curve by a pair of parametric equations, doing so arbitrarily 
without any thought of the geometric significance of the param- 
eter so introduced. Any pair will do if, when the parameter is 
eliminated, the original equation is produced. The parametric 
equations of the ellipse might have been introduced in this way ; 
for, given 

~2 "T 7.2 X J 



92 The Calculus. 

since 

cos 2 cf> + sin 2 </> = 1 

we are led at once to assume 



— = cos d> and -^- — sin <f>, 



as the parametric equations of the ellipse. 

In the same way, and for the same reason, we assume 

x = acos<f>] x=beo$<f> 

y = a sin (f> J y=bsm<f> 

as the parametric equations of the major and minor auxiliary 
circles respectively. It is evident from the figure that the param- 
eter cf> is the same in all three of these. 
Again, in 

x 2 y 2 

a 2 " b 2 ~ l > 



since 



if we assume 



then 



so that 



sec 2 <j> — tan 2 <j> = l, 

X 9 i 

— - = sec 2 <f>, 



i£-=tan 2 <£; 



x=asec<f>, y — b tan <f> 

are the parametric equations of the hyperbola. 

It is entirely unnecessary, in using this pair of equations in 
analytic work, to consider the geometric relations between <f> and 
the hyperbola. 

88. Derivatives in Connection with Parametric Equations. — 

All our derivative formulas can be applied directly to parametric 



Analytic Geometry. 93 

equations, but of course the parameter is the independent vari- 
able, so that the second ^-derivative of y cannot be formed by 
differentiating the differential of y and dividing by (dx) 2 , but 

must be formed by taking the ^-derivative of -p . This is best 

done by differentiating the first derivative and dividing by dx. 
As an illustration, consider the ellipse 

x = acos<f>, y = bsm<f>. 

Evidently, 

dx= — a sin <f> d<b] , , _ . _ — - 

, . ,! *\.ds=Va 2 sin 2 cf> + b 2 cos 2 4>>d<l>; 
dy — b cos <f> d<f> 

tanT= ^i=-^ C0t ^ 

,,_d(y>)_ ] q LcSc2 + ' d + 
dx -asm<f>-d<j>' 



or 



„ — b esc 3 <£ 

y — ^— ' 

P — — .// — 7 



a 2 
— r- (a 2 sin 2 d> + b 2 cos 2 <&)*. 



These results may be used as they stand, for to any given 
point of the ellipse corresponds a value of <j>, and substituting 
this value will give the slope, curvature, etc., at the given point. 
If desired, the results may be expressed independently of the 



1/ x 

parameter. Putting sin <f>= -j- 3 cos<f>=— , & 2 = a 2 (l — e 2 ), p 

o a 



be- 



94 



The Calculus. 



comes, for instance, 



P =^(a*-e^)K 



(Cf. Art. 84.) 
The value of p can be derived directly from the definition, 

ds 
p = -j- , and for some parametric equations this is much the 

more convenient way. 

Thus, since tan t = cot d>, 

a 

sec 2 t dr— — esc 2 d> dd>. 
a 



Substituting 1 + tan 2 t=1-\ — ^ c0 ^ 2 <£ ^ or sec2 T > we ^ n ^ ^ T anc * 

a 

ds 

get for t =p the same result as before. 

dr 

89. The Cycloid. — One of the most interesting of the transcen- 
dental curves is the cycloid, defined as follows: If a circle of 
fixed size rolls along a straight line, the curve described by a 
point of its circumference is called a cycloid. The line on which 
the circle rolls is called the directrix of the cycloid. The curve 
consists of an indefinite number of arches, touching the directrix 
in points called cusps; the piece of the directrix between two 

V_ 




Analytic Geometry. 95 

cusps is called the base of the corresponding arch, and the point 
of an arch furthest from the base is called a vertex. (See Fig. 
25.) 

If the radius of the rolling circle is a, the base of an arch is 
evidently 2ira, and the height 2a. 

Take the directrix as axis of x, and one of the points where the 
generating point touches the directrix as the origin; take the 
axis of y on the same side of the directrix as the rolling circle. 
Consider the position P(x, y) occupied by the generating point 
when the circle touches the directrix at M, its center having 
moved to 0', and the radius drawn to P having turned through 
the angle <j>. 

Evidently OM= arc PM=a<f>, and PN= a sin <f>; 

x=OM-PN. 
Also O'M—a, 0'N=a cos $; 

y = 0'M-0'N. 
Consequently 

x=a(cj> — sin<£),l 
y = a(l-coscf>).j 

are a pair of parametric equations of the cycloid. 

It is possible to eliminate cf> from these equations and express 
x as an explicit function of y, but the resulting form is incon- 
venient. From the parametric equations, however, the curve is 
readily studied. 

We can very easily obtain the coordinates of any number of 
points; for instance, with 

9 = u, -g- ; -^ ; -3-; it; -3-; -g-; -3-; ^^; etc., 
we have 

x = 0; .18a; .57a; 1.23a; ira; 5.06a; 5.71a; 6.10a; 6.28a , 

>-CtC 

y=0;.5a; a; 1.5a; 2a; 1.5a; a; .5a; 



96 



The Calculus. 



Since (tt— <£) and (v+<f>) give the same value to y, and give 
to x values differing equally from ira, the arch is symmetrical 
with regard to x — ira. Since increasing <j> by 2tt increases x by 
2ira and leaves y unchanged, the curve consists of equal arches, 
stretching indefinitely in both directions. 

Differentiating the equations, we have: 

dx—a(l — cos <£)d</> = 2&sin 2 ^- d<f>, 



dy — a sin <}> dcj> = 2a sin ■— cos 4r d<f>, 
ds 2 = dx 2 + dy 2 = ±a 2 sin 2 -|- (sin 2 ^ + cos 2 



W 



4a 2 sin 2 -^-(#) 2 , 



ds=±2a sin 



4> 



If 5 is measured from the origin and is positive in the direc- 
tion taken by the generating point as <f> increases, then, since 



d<j> 



is to be positive always, 



4> 



ds — 2a sin -^- d<j> for the arch from <f> — to cf> = 2ir; 
ds=—2a&m~-d<f> for the adjacent arches; and so alternately. 



dy 
dx 



tan t= cot -^- , 



dy . <2> 

-f- =sinr = cos -77- . 

eta . <£ 

-i — = cos r=sm -£-.: 
ds 2 



whence r— -Z -?- • 



Hence, in Fig. 25, the tangent at P is parallel to the bisector 



Analytic Geometry. 97 

of the angle PO'M, and consequently passes through the highest 
point of the rolling circle. 
The sub-normal is 

y tan r = 2a sin 2 -^- cot ~=asm <j>, 

and so =PN; the normal therefore passes through M, the lowest 
point of the rolling circle. 
The length of the normal is 

71 = y sec r = 2a sin 2 -~ esc -?-=2a sin ■—■ ; 

— z. z 

and the length of the radius of curvature is 

ds 2asiD 4^ . . 4, 



" «(W) 



(The positive value of p is — -, , since t decreases as s in- 
creases, r being measured from the positive direction of OX to 
the positive direction of the tangent.) 

Since p = 2n, the radius of curvature is bisected at the point 
M. 

90. As an example of the arbitrary use of the parameter, sup- 
pose we wish to study the curve given by 



(t) 



t /+*;-»■ 



If we letf- — J =cos 2 <f>, we findf 4- ) =sin 2 cf> ; whence 

x = acos 3 <f>, y = b$m 3 <f> 

are parametric equations of the curve. 

From these it appears that the .r-intercepts of the curve are 
± a, the ^/-intercepts ± b, and that these are the extreme values of 
8 



98 



The Calculus. 



x and y. As tan.T= tan <f>, the curve is tangent to each 

axis at its intercept, and so has four cusps. 

This curve, when b — a, is called the astroid. In this case, 

t=7t— <£, ds= ± 3a sin <j> cos <f> d<\>, + in the first and third quad- 
rants, — in the others, if s is 
measured positively contra-clock- 
wise around the curve. The 
length of the tangent is 

t = y cscr = y esc <£ = &sin 2 <f>. 

The distance along the tangent 
from its contact to the y-axis is 

t' — —x sec t — x sec (f> = a cos 2 </> ; 

hence t + t' = a; and the part of 
the tangent intercepted between 
the axes is of the constant length a. 

The length of the radius of curvature is 

p = ± 3a sin <j> cos <f> — : 




axy, 



3a 



with signs chosen as for ds. As p = -n~ sin 2<£, its greatest value 

f) 
is -^ , for the points half-way between two cusps. The normals 

at these points are -jr in length and reach to the origin. 

91. Examples. 

1. From the equations for the hyperbola, 
x = asec<f>, ?/ = &tan<£, 

find tan t, the subnormal, and the radius, of curvature. 

A . b b 2 (a 2 &m 2 cf> + b 2 )i 

Ans. tanr= — csc<£: s. n.= — sec d>- p =± — ^— — '— or 

a a r ah cos 3 </> 

(a 2 tan 2 4> + & 2 sec 2 <fr) 3 



Analytic Geometry. 99 

What limitations do these equations place on the value of ^ ? 

2. In the case of the ellipse 

x = acos<p, y=b&m<f>, 

given that the value <f> = <f> lt determines the point P x (x u y x ), 
show that <f> = cf> 1 ± \ determines the extremities of the diameter 
through the conjugate point. 

3. In the ellipse of example 2, show that if p is the radius of 
curvature at P x ; and 2r is the length of the diameter through the 
point conjugate to P 1? then ahp = r z . 

4. Show that x=at 2 , y — bt are parametric equations of a parab- 
ola; find in terms of t the lengths of the subtangent, the sub- 
normal, and the radius of curvature. Show that the extremities 

of the latus rectum are given by t= ± -~- , and find the values 

of the three lengths for these points and for the vertex. 

5. Show that a point at a distance b from the center of the 
rolling circle of the cycloid generates a curve having for its 
equations 

x=a<f> — b sin <j>, y=a—b cos <j>. 

(Any such curve is a trochoid.) 

Show that y"= &( flc p s <ft-&) . ^ that, if b<a, the points 
* (a-bcoscf>) 3 ' ' r 

given by <£ = cos _1 — are inflections. (Prolate cycloid.) 

Show that, if a<b, the values (j> x and —<f> 19 for which 

&m<f> J =^?- f give coincident points of the curve. (Curtate 

cycloid.) 

Show that for either trochoid, p = ± ^ — Jt — — 7 \ , 

r b(acoscp-b) 

and that for the points where the curtate cycloid cuts the direc- 
trix, p=Vb 2 — a 2 . 

6. Transform the equations of the cycloid, using for axes the 
tangent and normal at the vertex, y-axis upward, getting 

x' — a (<f>' + sin <£'), y' = a( — l + cos <f>). ($' = <{> — tt.) 



100 The Calculus. 

7. A jointed parallelogram PAOB has one vertex fixed at the 
origin of coordinates and moves so that the angle A OX is equal 
to twice the angle BOX; if the sides are OA = a, OB — b, show 
that the locus of P is given by the equations 

x=acos2<fx+b cos 0,1 
y=asm 2cf> + b sin <f>.j 

This curve is called a limagon. If b = 2a, it is the cardioid. 

Show that, for the cardioid, t=J(7t+30), p = -~- cos -?- • 
Find the values of x, y, r, and p for the points given by <f>=0, 
~o t "o~ j Kj -Q- 7 an( l -q- • Draw the curve. 



8. Find a pair of parametric equations for 

and show that they represent the part between (a, o) and (o, b) 
of the parabola represented by the four irrational equations in- 
cluded in 

*(tT* (I)'-- 

or by the single equation 

(JL + X_iY =4 iL. 

9. Find the parametric equations of the companion to the 
cycloid, the locus of the point N; and show that, if the origin is 
shifted to the point (na, a), the equations become 

x' — a<p f , y' — a cos $', 
where cp' = <f> — tt; so that the companion is the cosine curve 
x' 

w'rraCOS— . 

v a 

10. Show that 

sc = a(cos0+0sin0), y=za(sm<f> — <f> cos 0) 
represent a spiral for which p = a<j>. 

92. Polar .Coordinates. — We are familiar with the method of 
locating a point in a plane by giving its distances (called rec- 
tangular coordinates) from two given perpendicular lines. There 



Analytic Geometry. 



101 




^P(r&) 



P(-K0) 



are other methods of determining the position of a point in a 

plane, and of these the most important is by means of polar 

coordinates. If, as in Fig. 27, we choose a point of the plane 0, 

and a line, OA, through 0, then 

the distance r — OP and the angle 

6 = AOP are polar coordinates of 

P. The point is called the pole 

or origin, and the line OA the 

initial line of the system of polar 

coordinates. The distance r and 

the angle 6 are called the radius 

vector and the vectorial angle of 

the point P. The point is designated by {r, 6). 

The angle is measured counter-clockwise from the initial 
line if positive, clockwise if negative. The distance r is measured 
in the direction thus determined if r is positive, in the opposite 
direction if r is negative. 

93. Transformation of Coordinates. — The relation between 
polar and rectangular coordinates is simple if they have the same 
origin, and the same line as initial line and axis of x. Thus, in 
Fig. 28 we evidently have 



Fig. 27. 



x — r cos 



x- + y- = r 



y=r sin 6 f -^-=tan 





Fig. 28. 



Fig. 29. 



102 The Calculus. 

The equation of a curve in polar coordinates may be found 
either by expressing its geometric definition in terms of its polar 
coordinates or by transforming its equation as given in rectangu- 
lar coordinates. For instance, in the circle of Fig. 29, if the 
pole is on the circumference, and the initial line is the diam- 
eter through 0, it is evident from the geometry of the figure that 

r=2a cos 0, 
a being the radius. 

Again, the equation of this circle in rectangular coordinates is 

(x — a) 2 -\-y 2 = a 2 , or x 2 + y 2 = 2ax, 

which, through the formulas of transformation, becomes in polar 
coordinates 

r 2 — 2ar cos 0, 
or 

r=2acos 6. 

As another example, consider the parametric equations 

x — a cos 2<j> + 2d cos <f> "J 
y = a sin 2<f> + 2a sin <j> J 

which represent (Ex. 7, Art. 91) a cardioid. If we first shift the 
origin a distance a to the left, these become 



x=a+acos 2<{> + 2a cos <£, 1 
;in <f>. \ 



y=a sin 2<£ + 2a sin 

These, since 1 + cos 2cf> = 2 cos 2 <f>, sin 2<j> = 2 sin <f> cos<£, may be 
written 

x=2acos <£(l + cos<£) 
y=2asm </>(l + cos $) 
Evidently 

tan0=-^- = tan<£; = 0. 



■•} 



r 2 = ^ 2 + i/ 2 =:[2a(l + cos0)] 2 (sin 2 </> + cos 2 0), 
r = 2a(l-fcos </>) ; 



Analytic Geometry. 103 

and r = 2a(l + cos0) is the polar equation of the cardioid, the 
pole being the cusp, the initial line the axis of symmetry. 

From this equation, r = 2acos0 + 2a, it is evident that the 
cardioid is the locus of points distant 2a further from the origin 
than the points of the circle r—2a cos 6. 

94. Examples. 

Find the polar equations of the following curves, taking the 
origin and a>axis of the rectangular system as the pole and 
initial line of the polar system, except when directed otherwise. 

1. x = a; y = b. 

2. x 2 + y 2 = a 2 . 

3. Ellipse : major axis 2a, eccentricity e, pole at left focus, 
initial line the major axis. (Hint : After substituting, solve the 
equation for r. ) 

Ans . r= "(l-* 2 ) or r= _ a(l-e-) 
1 — e cos 6 1 + e cos 9 

4. Parabola : parameter p, pole at focus. 

Ans. r — p esc 2 f or r= — p sec 2 f . 

5. The hmniscate (Art. 74), (x 2 + y 2 ) 2 = a 2 (x 2 -y 2 ). 

Ans. r 2 = a 2 cos 20. 

6. The cardioid (Art. 74), (x 2 + y 2 ) 2 + 2ax(x 2 + f) -a 2 y 2 = 0. 

Ans. r—— a(l + cos#) or r=a(l — cos 0). 
In each of the examples with two answers, tracing the curve 
from the two polar equations will show that the two equations 
are equivalent. This may also be seen from the fact that the 
point determined from one equation by 6=a is the same as the 
point determined from the other by 0=a + 7r. 

95. Derivatives with Polar Coordinates. — Let a curve be given 
by an equation in polar coordinates, and let it be required to find 
the angle if/ from the positive direction of the radius vector at 
any point P to the tangent at the same point. (See Figs. 30 
and 31.) This value can be obtained directly, or by transform- 
ing to polar coordinates the results already obtained in rectangu- 
lar coordinates. Following the second method we have : 

x=rcos6, dx = cos0 dr— r sin dO 
y=rsm$, dy — §mO dr + r cos Odd 



;•} 



104 



The Calculus. 



tan t— 



dy^ 
dx 



sin dr + cos 9- rdO 



cos dr — sin 6 - rdB ' 
Dividing each term of this fraction by cos $ dr, we have 



tan 



tan#+ ^=- 
dr 



l-tan0 



rdO 
dr 



But since t=6 + \J/, 



Hence 



, _ tanfl+tani/r 
~ 1 — tan tan \J/ 



tan^=-^- 



dr 



As r=f(d), -Tn —f (#) is represented by /; so 



d$ 



tan^= —r 



The differential of arc ds is obtained similarly : 

(ds) 2 = (dx) 2 + (dy) 2 = (rd$) 2 + (dr) 2 . 
The other functions of \p may thus be expressed : 
2 , 1,4. 2 , (rd0) 2 +(dr) 2 (ds) 2 

dr 



etc. 



COS tf/: 



. , rdO 



etc. 




Fig. 30. 



96. These results are conveniently 
viewed in the following figure, called 
the polar differential triangle. 

In Fig. 30, regarding the arcs PP' 
B and P'B as straight lines of lengths 
ds and rd6, PB as of length dr, B as a 
rectilinear right angle, and P'PB as 
the angle \f/ between the radius vector 
OP and the curve, the results just ob- 
tained are easily remembered: 



Analytic Geometry. 



105 



tan \p — -3- : 
r ar 9 



CO&xp- 



dr 
ds 



Sin \p: 



rdO 
ds 



It is usual to regard as the independent variable in all 
treatments by polar coordinates. 

97. Applications of the Polar Differential Triangle. — Cor- 
responding to a point P(r,6) JV 
of a curve r=f(6) (Fig. 31), 
draw the radius vector OP, 
the tangent PT, the normal 
PN, and a perpendicular to 
OP through meeting PT 
and PN at T and #. 

PT and PN are called the 
polar tangent and normal, OT 
and ON the polar subtangent 
and subnormal. Let p be the 
perpendicular from to PT. Fig. 31. 




Pr=tangent =r sec f= *£ _r ^1+ (-^-)* - £ VF+?* 



r 2J^ ~2 

0!T= subtangent =r tan ^= -p = — 



OiV= subnormal =rcot^= ^-~ = -~=r r . 



p=rsin^r = 



r 2 ^ 



ds V(dr) 2 +(nZ0) 



106 The Calculus. 



P I 2 . / dr V Vt^+V 2 ' 



i-+m 



ds 

The radius of curvature is found from the definition p = -5- , 

dr 



and the relations t=ij/ + 0, tan ^= -^- 3 whence 

■ _ r+/ tanfl _ r cos fl + / sin 
~ r' — r tan 9 ~~ r' cos — r sin * 

98. Examples. 

1. Find for r=2acos#, the six lengths of Art. 97. 

2. Show that for the cardioid r = 2a sin 2 J, 

T=-rt- ±w*r, ^=2asin|<i0 ; /o = §V2a.r. 

3. Show that for the lemniscate r 2 = a 2 cos 20, 
rr , «/, r 3 a 2 



*=-g+8», p = ^, P = - 



4. Show that in the parabola,, r=asee 2 f , Vr 2 -i-/ 2 =asec 3 |. 
the polar subtangent=2acsc 0, and the perpendicular from the 
focus on a tangent = a sec |. 

5. Find the radius of curvature of the parabola r=asec 2 |, 
and its values for the vertex and the extremities of the latus 
rectum. Ans. /o = 2asec 3 f ; 2a, and 4a V2. 

6. Find the radius of curvature of the ellipse r= ^ ^ , 

r 1 + e cos 

and its values for the points given by 0=0, \, tt, cos _1 ( — e). 

7. Find tan^ for the Spiral of Archimedes, r=a$, and for 
the Logarithmic or Equiangular Spiral r = ae n9 . 

8. Show that the radius of curvature for r = ae n9 is rVl + n 2 , 

(1 4-0 2 )§ 

and for r = aO is v , J a. 

a -f-U 

Higher Degree Curves. 
99. Curve-Tracing. — In order to find the form of a curve from 
its equation, we can either plot the curve or trace it. In plot- 
ting a curve, we find the coordinates of a large number of points, 



Analytic Geometry. 107 

and by marking these points obtain a dotted line to serve as a 
guide in sketching the curve. In tracing a curve, we locate a 
few important points, find approximate forms of the curve at 
these points, and then sketch the curve. Thus an ellipse can be 
constructed point by point with the aid of the auxiliary circles; 
or a circumscribed parallelogram may be used as an approximate 
form, as in the algebra ; or again, the circles of curvature at the 
ends of the axes will furnish closer approximations. 

100. Approximate Forms. — If f(x, y)=0 is the given curve, 
and <f>(x, y) =0 is another, meeting f(x, y) =0 in two or more 
points coincident at P, <f>(x, i/)=0 is a tangent or an approxi- 
mate form to f{x, y) = at P. The greater the number of points 
coincident at P, the closer is the approximation. 

Approximate Forms at the Origin and at Infinity. — When two 
equations are solved simultaneously, the number of common 
solutions is in general the product of the degrees of the equa- 
tions. It is always easy to see how many of these solutions have 
the value zero or are infinite. 

For instance, consider the cubic equation 

x 3 + x 2 -y = 0. (1) 

It should have three solutions in common with the linear equa- 
tion 

y=0. (2) 

But x z + x 2 = or x 2 (x+l) = gives x = twice; so (0, 0) is 
a double pair of solutions, and the z-axis is tangent to the curve 

(i). 

The curve y = x 2 should have six intersections with (1) ; but 
x 3 = is of the third degree only, and, considered as of the sixth 
degree, has three zero roots and three infinite roots. Hence the 
parabola y — x 2 meets (1) three times at the origin and three 
times at infinity. (See Art. 46, Algebra.) The parabola y—x 2 
is a closer approximation to (1) at the origin than the tangent. 



108 



The Calculus. 



The curve y=x s should meet the curve nine times; but x 2 = 0, 
considered as an equation of the ninth degree, has two roots cor- 
responding to the origin, and seven infinite, y—x 3 is therefore 
a closer approximation than y — x 2 at infinity, but not so close at 
the origin. 




Fig. 32. 



We therefore take y=x 2 as the form at the origin, and y—x 3 
as the form at infinity, using the part of y — x 2 near the origin 
and the remoter parts of y — x 3 . (See Fig. 32.) 

From the foregoing, the significance of the following general 
principles will be clear. 

The terms of lowest degree give the form at the origin (if the 
curve goes through the origin) ; the terms of highest degree give 
the form at infinity. 



Analytic Geometry. 



109 



The terms of the very lowest degree give tangent lines at the 
origin ; terms not homogeneous, but lower in degree than the rest 
of the equation, give a curvilinear form at the origin. 

The terms of the very highest degree give lines intersecting the 
curve at infinity, but not necessarily tangent to it; terms not 
homogeneous, but higher in degree than the rest of the equation, 
give a curvilinear form, always tangent at infinity, but not neces- 
sarily of as close tangency as possible. 

101. Asymptotes. — Tangents to a curve at infinity are called 
asymptotes. In Art. 128 of the Algebra, a method was given of 




Fig. 33. 



determining the asymptotes of a hyperbola. This method applies 
equally well to curves of higher degree, both in demonstration 
and in application. Thus any real factor of the terms of highest 
degree in the equation may be evaluated to give an asymptote. 
If there are no such real factors, there are no infinite branches, 
and the curve is closed. If the evaluation gives an infinite result, 
the corresponding branch is parabolic. 
The curve 

x 3 -\-y 3 — 3xy= 



110 



The Calculus. 



has for tangents at the origin 

xy = or x=0 and y=0, 
the coordinate axes. Also, 



x + y- 



3xy 



x 2 -xy + y] y= _ x= 



Sx 



3z 2 



l-u 



so that x + y=—l is an asymptote of the curve. (See Fig. 33.) 
Again, 

2x 2 y + y 2 + 8x = 




Fig. 34. 

has for tangent at the origin 8a: =0,' the y-axis, and for a closer 
approximation at the origin 

f=-8x, 

a parabola. 

Evaluating y(2x 2 + y) = -8x for the parabolic factor, we get 

-8x1 4 

2x 2 + y = 



y 



f=— 2x 2 =os 



o, 



or the parabola y——1x % as a curvilinear asymptote. 



Analytic Geometry. 



Ill 



Evaluating for the linear factor, we get 



-&c ] A_ n 

or the ar-axis, y=0, as a rectilinear asymptote. (See Fig. 34.) 



102. Typical Forms at the Origin and at Infinity. — The para- 
bolic forms y — x 2 , y — x 3 , y 2 = x 3 , etc., of the general form 
ym — ± x n ^ are f frequent occurrence as approximate forms ; each 
of them is readily traced by using the general principles above. 
For instance, in y—x 2 , the term of lowest degree gives y = 0, the 



Y 




y = r 



y=x* y 2 = x 3 

Fig. 35. 

rr-axis, as tangent at the origin, and y is evidently positive for 
any real value of x. (Fig. 35.) 

y — x 3 is also tangent to the rr-axis at the origin, and y is of 
the same sign as x. 

y 2 — x 3 is tangent to the z-axis at the origin, and x is positive 
for all real values of y. 

x 2 — — y 3 is tangent to x = 0, the y-axis, and y is always nega- 
tive. This is the same curve as y 2 = x 3 , but with the cusp point- 
ing upward. 

A second-degree parabola of the form y 2 — ax has a/2 for the 
radius of curvature at the vertex ; this enables us to approximate 



112 



The Calculus. 



more closely to the proper curvature at the origin for a curve 
having y 2 — ax for its lowest terms. For instance, the curve of 
Fig. 34, having the form y 2 = — Sx at the origin,, is very closely 
tangent to a circle of radius 4 with its center on the #-axis at 
(-4,0). 

The forms at infinity of the parabolic types, y=x 2 , y — x z , 
y 2 = x 3 , etc., are from their highest terms similar to x=0, or the 
y-Sixis. As a point (x, y) moves out on one of these curves, 
x and y both increase indefinitely, but the curve becomes more 
and more closely parallel to the ?/-axis. In the same way, x — y 2 , 
x=y s , x 2 = y s , etc., go further and further from the axes, becom- 
ing more and more nearly parallel to the a>axis. 



xy=l 



x 2 y = l 
Fig. 36. 



W 



The hyperbolic forms, xy—\, x 2 y — l, xy 2 — l, etc., of the gen- 
eral form x 1l y m =±l, evidently have the coordinate axes as 
asymptotes, since for each y=oo when x = and x= oo when 
2/ = 0. For xy = l, x and y are of the same sign; so the two 
branches are in the first and third quadrants. For x 2 y — l, y is 
positive for all real values of x; so the two branches are in the 
first and second quadrants; xy 2 — \ is in the first and fourth 
quadrants, xy— — 1 is in the second and fourth, and so on. 

103. Uses of Derivatives. — Besides the forms at the origin 
and at infinity, it is often necessary to find the form of the curve 



Analytic Geometry. 113 

at some other important points. Generally no more is done than 
to find the slope at some convenient points, especially the inter- 
sections with the axes, and to locate the points where the curve 
has some particular slope, especially Oor oo, sometimes ±1. 
For instance, the curve of Fig. 32, 

y=x 2 + x 3 , 

cuts the £-axis at ( — 1, 0). Its slope is in general 

4f L =2x+3x 2 , 
ax 

and is +1 at this point. Again, its slope becomes when x=0 
or — § ; so the curve is parallel to the a>axis at ( — § , ^ 4 T ) . These 
data aid materially in giving the proper form to the curve. 
Again, the curve of Fig. 34, 

2x 2 y + y 2 + $x=0, (1) 

has for its slope 

dy 4(sy+2) 

dx 2(x 2 + y) ' 

and this slope is for the point on (1) for which xy— — 2. 
Putting y= —2/x in (1), 

-4z+-4- +8x = 0; x=-l, y=- — =2. 
x 2 v x 

Thus the curve is parallel to the a>axis at ( — 1, 2) . 

The slope of (1) is oo for the point on (1) for which y=—x 2 , 
or 

-2x 4 + x* + 8x = 0, 
whence 

x=0 or 2 and y=— x 2 = or —4. 

Thus the curve is parallel to the ?/-axis at the origin and at 
(2, -4). 
The form of the equation (Fig. 33) 

x s + f-3xy=0 
9 



114 



The Calculus. 



shows plainly enough that the curve is symmetrical with regard 
to y — x } so that the end of the loop is its intersection with y—x. 
The tangents at the inflections of a curve are of great assist- 
ance in drawing the curve, and are sometimes readily found, par- 
ticularly if the equation can he put in the form y=f(x). 

104. Analysis of the Equation. — The general principles given 
for selecting terms out of an equation to represent a valuable 
approximation to the curve at the origin or at infinity are very 
broad in application, but do not cover all cases. 

For instance, in x 3 + y 3 — 3xy — 0, 

x 3 — Sxy=x(x 2 — 3y) =0 and y* — 3xy=y(y 2 — 3x)=0 

are of the same degree as the curve, but give two parabolic forms 
at the origin, as well as the tangent axes. It is a tiresome process 
to test all the combinations of terms in an equation in order to 
find the ones which give close approximations, but the selection 
of such terms is greatly simplified by the employment of a 
graphic analyzer called the analytical triangle. 

105. The Analytical Triangle. — The analytical triangle is 
formed by plotting each term of a complete equation, in rectan- 
gular coordinates, with the exponents of x and y for the co- 
ordinates of the corresponding point. The analytical triangle 

A OB of the cubic equation is 
given in Fig. 37; a complete 
equation of the n-ih. degree 
would have (n + 1) points on 
each side of the triangle form- 
ed by joining its outside terms. 
The sides OA and OB are 
called the analytical axes of x 
and y, respectively. The points 
— on the analytical axis of x rep- 
resent all the terms of the 
equation which do not contain 



y 


B 


f 


% \. 


y 


1 -X 

1 l\ 




! i \4 





«A» Hi jC 



Fig. 37. 



Analytic Geometry. 



115 



y; the points on the analytical axis of y, the terms which do not 
contain x; while the points on the side AB represent all the 
highest degree terms of the equation, and, as we have seen, 
determine the form of the curve at infinity. 

The vertices 0, A, and B are called the fundamental points 
of the analytical triangle; the vertex corresponds to the origin 
of coordinates, the vertex A to the point at infinity in the direc- 
tion of the axis of x, and the vertex B to the point at infinity in 
the direction of the axis of y. 

106. The equation of a given curve is placed upon the analyti- 
cal triangle, when all the terms which compose it are designated 
by some distinguishing mark. The cubic equation 

x 3 + y 3 -3xy + 2x 2 + y = (1) 

is placed upon the analytical tri- 
angle in Fig. 38, where each term 
is indicated by drawing a small 
circle around it; no attention is 
paid to its coefficient, as the ob- 
ject of the graphic analysis is to 
indicate the terms which give the 
equations of approximate forms, 
and the meaning of the missing 
terms of the complete equation. 

107. The Analytical Polygon. — If straight lines are drawn 
joining marked points of the analytical triangle so as to form a 
convex polygon, outside of which no marked point lies, the figure 
thus formed is called the analytical polygon. The polygon abed 
in Fig. 38 is the analytical polygon for the equation given in 
Art. 106, the marked point within it having no significance in 
determining the general characteristics of the curve. 

The analytical polygon must have either a side or a vertex 
on each side of the analytical triangle. An equation formed 
by setting equal to zero the terms of the original equation 




116 The Calculus. 

upon any side of the polygon, enables us to obtain an ap- 
proximation to some part of the curve. For brevity's sake, we 
say that the side of the polygon gives the form of this part of the 
curve. An outer side of the analytical polygon is one such that 
every marked point lies between it and the origin; an inner side 
is one such that no marked point lies between it and the origin; 
the remaining sides, if any exist, lie on the legs (analytical axes) 
of the analytical triangle. 

(1) An outer side gives an approximate form of the curve at 
infinity, indicating always a linear or a curvilinear asymptote. 

(2) An inner side gives an approximate form at the origin. 

(3) A side on one of the analytical axes gives, by solution, the 
intersections of the corresponding coordinate axis with the curve. 

While these three precepts furnish the fundamental analysis 
of an equation by means of its analytical polygon formed by the 
marked points of the analytical triangle, the unmarked points 
are of importance in determining the curve, especially if they 
include one or more of the fundamental points. 

If the point is unmarked, the equation has no absolute term, 
and the curve passes through the origin. If one of the points 
A or B is unmarked, it is at the vertex of a small triangle cut 
off from the analytical triangle by an adjacent side- of the ana- 
lytical polygon, or by such a side produced; this side is said to 
cut off the unmarked vertex, and gives the approximate form of 
the curve only in the direction corresponding to the unmarked 
vertex. If the terms forming such a side contain a monomial 
factor, it is to be disregarded; if such factor should give an 
asymptote, this may be found more advantageously from some 
other side of the polygon. 

The terms of a side lying on AB, being homogeneous, may 
always be decomposed into factors of the first degree, either real 
or imaginary ; each of the real linear factors gives a direction in 
which the curve goes off to infinity, and by evaluating such fac- 
tors by the method employed in Art. 101, we may determine the 



Analytic Geometry. 117 

asymptotes if any exist ; if the evaluation of a linear factor does 
not lead to a finite value, the curve has no asymptote correspond- 
ing to it, but goes off to infinity in a parabolic form. It is not 
necessary to evaluate the monomial factors, as the infinite 
branches corresponding to them are given to better advantage by 
the sides of the analytical polygon cutting off the unmarked 
vertex A or B. 

108. The sides of the analytical polygon abed in Fig. 38 will 
furnish the following analysis of the given equation : 

x s + y 3 -3xy + 2x 2 + y = 0. 

The outer side ab gives 

x 3 + if = (1) 

as the first approximation of the infinite branch, showing the 
linear asymptote corresponding to the factor (x+y). 

The side be, lying in the analytical axis of y 3 gives the equation 

i/ + y=0, (2) 

which determines the intersections of the given curve with the 
axis of y. 

The side ad, lying in the analytical axis of x, gives 

x 3 + 2x 2 = 0, (3) 

which determines the intersections of the curve with the axis 
of x. 

The side cd, since the analytical origin is unmarked, gives the 
approximate form of the curve at the origin, 

2x 2 + y = 0. (4) 

Equation (2) shows the curve to intersect the axis of y in but 
one real point, the origin. 

Equation (3) shows the curve to intersect the axis of x twice 
at the origin and at ( — 2, 0). 



118 



The Calculus. 



The slope, 



dy -3x 2 + 3y-4:X 
dx ~ 3y 2 -3x + l 



-4 



, is —at (-2,0). 



Equation (1), when evaluated from the other terms of the 
equation, gives the asymptote 



/ , \_3xy — 2x 2 —y 



y=— x=<x> 



5x 2 

3x 2 



(5) 




Fig, 39. 



The curve is now readify traced, by drawing the asymptote, 
drawing in the form at the origin, and marking the only other 
point where the curve crosses the axes. (See Fig. 39.) 

109. The three equations which we have already discussed in 
Arts. 100 and 101 we shall now subject to more complete analysis 
by reference to their analytical polygons. 

For y = x 2 + x* (Figs. 40 and 32) : 

y = gives one intersection with the y-axis at the origin, two at 
infinity. 



Analytic Geometry. 



119 



x 2 + x s = gives two intersections with the a>axis at the origin, 
and one at ( — 1, 0). 

y = x 2 is the form at the origin. 

y — x 3 is the form at infinity in the direction of the ?/-axis. 

For x s + y 3 -3xy = (Figs. 41 and 33) : 

y z = gives three intersections with the axis of y at the origin, 
and x 3 — gives three with the axis of x. 

x 3 -\-tf must be factored and its real factor evaluated, giving 
x + y= — 1 as the rectilinear asymptote. 




Fig. 40. 



Fig. 41. 



Fig. 42. 



The forms at the origin are : 

a' 3 — oxy = x(x 2 — 3y)=0 
and 

y 3 -3xy=y(tf-dx)=0, 

a parabola x 2 = 3y tangent to the .r-axis and above it, and a parab- 
ola if — 3x tangent to the y-axis and to the right of it. 

For 2x 2 y+y 2 + 8x=0 (Figs. 42 and 34) : 

y 2 = gives two intersections with the y-axis at the origin, one 
at infinity. 

x = gives one intersection with the .r-axis at the origin, two 
at infinity. 

y 2 —— Sx is the form at the origin. 

2.r 2 y + y 2 = or y=—2x 2 is the form at infinity in the direc- 
tion of the axis of y. 



120 The Calculus. 

2x 2 y + 8x=0 or xy — — 4 is the form at infinity in the direc- 
tion of the axis of x. (The curve bears no resemblance to 
xy— — 4 in the other direction.) The curve therefore approaches 
the a>axis as an asymptote, lying below it on the right, and above 
it on the left. 

110. Parametric Equations x—f{m), y=mf(m). — If a curve 

has two or more tangents at the origin, that is, if the lowest 
terms of its equation are of at least the second degree, points 
of the curve may be determined by a certain pair of parametric 
equations much more readily than by the single equation. These 
parametric equations are determined by assuming 

y — mx, 

and thence expressing both x and y as functions of the parameter 
m. For instance, applying this method to the strophoid, 

x(x 2 + y 2 )+a(x 2 -y 2 )=0, 
we find 

m 2 — 1 m 2 — 1 

— o , -. a. y = m 2 ■ -, a, 
ra 2 -fl ' u m 2 -rl 



x — 



from which, by assuming various values of m, any number of 
points of the curve may be determined. 

The parameter m of any point of a curve is of course the slope 
of the radius vector to the point, or is the same as tan 6 in polar 
coordinates. 

111. Examples. 

1. y 2 (2a—x)=x 3 . (Cissoid.) 

2. (a 2 -x 2 )y 2 = a 2 x 2 . 

3. (a 2 + x 2 )y 2 = a 2 x 2 . 

4. y(a 2 + x 2 )=a s . 

5. (fy 2 =(a 2 -x 2 )\ 

6. a?y 2 =\a 2 -x 2 )x\ 

7. x 2 (a 2 -y 2 )=a\ 

8. y 2 (a 2 -x 2 )=x\ 



Axalytic Geometry. 121 



9. axy—x 3 = 2a 3 . 

10. a 2 y-U 2 x + x 3 = 0. 

11. y 2 (x-2a)+U 2 x = 0. (Witch.) 

12. xy 2 = a(x 2 + a 2 ). 

13. a 3 y=(a + x) 2 (a 2 -x 2 ). 

14. y 2 (a — x)=x 2 (a + x). (Strophoid.) 

15. a?-y*-x + y 2 = 0. 

16. y*-3ax2f + 2ax 3 = 0. 

17. y 3 + ax 2 — axy — 0. 

18. a; 4 + 2ai/ 3 -3aa:?/ 2 r=0. 

19. x* + y* = a 2 xy. 

20. z 4 + 2aV-7a 2 z?/ + 3ay = 0. 



CHAPTEE III. 
Maxima and Minima. 

112. Maxima and Minima, Extrema. — Sometimes the graph 
of a function has a point which is further from the axis of x 
than any other point of the graph in the immediate vicinity, as at 
A or B in Fig. 43. The value of the function at such a point 




is an extremum; a maximum if it is larger than the values near 
hy, as at A, a minimum if it is smaller, as at B. 

If the coordinates of A or B are (x , y ), y is an extreme 
value of the function, x is the value of the independent variable 
that makes the function a maximum or a minimum, and 

[*U-M« -"«>-•■ 

This principle ~is not altogether general, for the graph may 



Maxima and Minima. 123 

have cusps, as in Fig. 44, where 
f(x) is an extremum, although Y 

f (x) is not 0. These cases, how- _ A 

ever, we shall disregard. 

With simple functions, the con- 
verse of the principle is commonly 
true (i. e., f(x) has an extremum FlG 44 

whenever f(x)=0), but it is evi- 
dent that if the graph has an inflection where f(x)=0, the 
function has not an extremum (C, Fig. 43). 

On either side of a point where f(x) is a maximum, 
taLTiT = f(x) is decreasing as x increases, and its derivative, 
f"{x) % is negative. On either side of a point where 'f(x) is a 
minimum, tanT = /'(a;) is increasing as x increases, and its de- 
rivative, f'(x), is positive. 

We therefore have as tests : 

f(x)=0, f'(.r)<0, f(x) is a maximum. 
f(x)=0, /"(#)>(), f(x) is a minimum. 

The cases in which f(x)=0 and f'(x)=0 include all such 
inflections as C in Fig. 43, and also points at which f(x) has an 
extremum (e. g., y — x^ when x = 0). 

The test for extrema may be stated as follows : 

If, as x increases through the value x , f(x) passes through 
the value zero and changes sign, f(x ) is an extremum; if f(x) 
changes from + to — , a maximum; if from — to +, a 
minimum. 

In geometric problems, it is almost always easy to see whether 
an extremum exists or not, and of which kind it is. 

If y — f(x) is positive, y 2 and y have extreme values of the 
same sort for the same values of x; for the sign of dy 2 = 2ydy is 
the same as that of dy, and both become zero together. Again, 

y and — have extreme values of opposite sorts for the same value 



124 The Calculus. 

of x; for d — = \ is opposite in sign to dy and becomes zero 

when dy = 0. We may thus often simplify the problem of finding 
the extreme values of a given function, by finding those of the 
square of the function, or of its reciprocal, and interpreting the 
results accordingly. 



113. Examples. 

1. Find the dimensions of the open box of greatest capacity 
which can be made from a piece of tin plate 3 in. square, by cut- 
ting a small square from each corner and folding up the edges. 

Let #:=the side of the small square; i. e., the depth of the box. 
Then (3 — 2x) is a side of the bottom, and the volume V is 

V = x(3-2x) 2 . 
^V=V f = 3(3-2x)(l-2x)=--0 when z=for ^. The ex- 
treme values of V therefore occur when x — f and when x—\, 
and are F = 0, the minimum, and V — 2 cu. in., the maximum. 

2. To find the cylinder of revolution of given volume that 
shall have the minimum surface (for instance, a closed tin can 
of given volume, of such dimensions as to require the least 
amount of material in its construction). 

Let y = the height of the cylinder and x — the radius of the 
base; then 8 = 2ttx 2 + 2Ttxy , where x and y are subject to the re- 
strictions that the volume shall be constant, or 

V = Trx 2 y=C. 

By elimination, 8 might be expressed as an explicit function 

of x or y, and the value of x or y found to make-j— = 0, or 

— - T — — 0. It is simpler however to proceed as follows : 
dy 

d8 = 27r] {2x + y)dx + xdy\, 

which will be zero when 8 is a minimum. 
Since V is constant, we also have 

dV = 7r(x 2 dy-{-2xydx) =0, 



Maxima and Minima. 125 

and from this equation 

dy — *- dx. 

x 

Substituting this value of dy in the value of dS, 

dS = 27r\ (2x + y)dx-2ydx\, 
and 

-g=M3*-y)=o, 

if y=2x. 

We thus have the proper proportions of the cylinder, and its 
actual dimensions can be computed from the relations 

V=7rx 2 y and y — 2x, 
from which 



V = 2ttx 3 and x 



-Ml 



3. Prove that the maximum area of a rectangle which can be 

X 77 

inscribed in the ellipse — ^ + 4* = 1, having its sides parallel to 

the principal axes, is 2ab. 

4. Prove the results of example 3, using the parametric equa- 
tions x—a, cos <j>, y — b sin <f>. 

5. Prove that the maximum volume formed by revolving a 
rectangle inscribed as in examples 3 and 4 about the z-axis is 

4^rb 2 a 
3V3 ' 

6. A triangle is inscribed in a parabolic segment having a base 
2b and altitude a, the vertex of the triangle being at the mid- 
point of the base of the segment and the base of the triangle 
parallel to the base of the segment; find the maximum value of 

the area of the triangle. Ans. A= Q /ir . 

7. An open cylindrical can is to contain 231 cu. in. What is 
the least amount of tin that can be used to make it ? 

Ans. 165.42 sq. in. 



126 The Calculus. 

8. Find the most economical proportions for a closed cylin- 
drical tin can, if in making each of the circular ends it is neces- 
sary to use up a piece in the shape of a regular hexagon circum- 
scribing the circle. 

Ans. h being the height and D the diameter, i = 2 ^ 3 = |1 

D 77 10 

nearly. 

9. Find the most economical proportions for a cylindrical tin 
cup, making the same allowance for waste as in the preceding 

example. Ans. -^ = _ = - F — nearly. 

10. A man is in a rowboat, 4 mi. from the nearest point, A, 
on a straight beach, and is bound for a point B on the beach, 25 
mi. beyond A. He can row 2^ m/h, and walk 3 J m/h. Where 
shall he land to reach his destination as soon as possible ? 

Ans. |V~6 mi. from A. 

11. The altitude of a right circular cone is 1%, and the radius 
of its base is a. Find the greatest volume of an inscribed right 
circular cylinder. Ans. V—-^ T 7ra 2 h. 

12. Find the dimensions of the cylinder in example 11 if its 
total surface is to be a maximum. 

Ans. Eadius = ** , height = ffi" 2 ^. . 
2(h — a) 2(h—a) 

13. Determine the cone of maximum lateral surface and the 
one of minimum lateral surface inscribed in a paraboloid of 
revolution of height a, and diameter of base a. 

Ans. The altitudes are — and -q respectively. 

14. In a circle of fixed radius a a rectangle of sides 2x and 2y 
is inscribed; the figure is revolved about the diameter perpen- 
dicular to the side 2x. Find the dimensions so that (a) the 
rectangle shall have the maximum area, (b) the cylinder shall 
have the maximum volume, (c) the cylinder shall have the 
maximum lateral surface. 

Ans. (a) and (c),x=y=-j^; (b) «=-|- VS, y= -S=. 



Maxima axd Minima. 12 7 

15. From a steamer A, going north at 8 m/h, a steamer B is 
observed going west at 10 m/h. If A turns just as B crosses its 
path, what straight course must A take in order to cross the 
course of B as near B as possible? 

Ans. About N. 53° 8' W. 

16. The base of a column 9 ft. high is 16 ft. above the eye of 
an observer. How far off must he stand for the column to sub- 
tend the greatest possible angle? Ans. 20 ft. 

IT. An isosceles triangle is circumscribed about a parabolic 
segment, the base of which is parallel to the tangent at the ver- 
tex. Show that the area of the triangle is least when its altitude 
is |- of the altitude of the segment. 

18. The expenditure of coal in steaming a ship is proportional 
to the time and to the cube of the speed ; find the most econom- 
ical speed against a current having the speed a. 

Ans. -=- actual speed, -^ through the water. 

19. A circular sector is to have a given perimeter and as large 
an area as possible; what must be its angle? 

Ans. 2 radians. 

20. The strength of a beam is proportional to the breadth 
and the square of the depth. Find the dimensions of the strong- 
est beam that can be cut from a cylindrical log of radius a. 

Ans. |V3by^V6. 

21. A triangle is inscribed in an ellipse, its vertex at a vertex 
of the ellipse, its base a double ordinate. Find the greatest area 

it can have. Ans. — — Vo, a and b bein^ the semi-axes. 

4 c 

22. If the figure of example 21 is revolved to form a cone 
inscribed in an ellipsoid, what is the greatest volume the cone 
can have? Ans. ff -n-ab 2 . 

23. A rectangular strip of copper is to be bent so that its 
cross-section is a circular arc; show that to give the maximum 
capacity, the arc must be a semicircle. 

24. Two circular plates, each of radius a, are to be cut and bent 
into conical surfaces and put together to form a can-buoy. What 
must be the radius of the base of each cone if the buoy is to be as 

large as possible? Ans. -r- V6*. 



128 



The Calculus. 



25. A wall 27 ft. high is 64 ft. from a house. Find the length 
of the shortest ladder that will reach the house if one end rests 
on the ground outside of the wall. Ans. 125 ft. 

26. Find the least volume that can be left between a sphere 
and a circumscribed cone of revolution. (Hint: Find two ex- 
pressions for the area of the section of the cone through its alti- 
tude, and thus get a relation involving x, y, and a, the radius of 
the base and the altitude of the cone, and the radius of the 
sphere.) Ans. frf. 







CHAPTER IV. 
Integration. 

114. Definition. — If f(x)dx is the differential of F(x), F(x) 
is an integral of f(x)dx. 

For instance, since 2xdx is the differential of x 2 , x 2 is an in- 
tegral of 2xdx. d(x 2 + c), if c is any constant, is also 2xdx, and 
in general : 

F(x) being an integral of f(x)> dx, and c being any constant, 
F(x) +c is also an integral of f(x)dx; that is, any differential 
expression has innumerable integrals, any two of which differ by 
a constant. N 

Thus integration, as the process of finding an integral is called, 
is the inverse of differentiation, and like most inverse processes, 
leads to multiple-valued results. 

The notation used for integrals and for integration is the 
following : 

$f(x)dx = F(x)+C; this is read: "The integral of f(x)dx 
is F(x) plus some constant." The constant C is spoken of as an 
arbitrary constant, because it may be any constant whatever; it 
must be written if the relation above is to be used as an equation. 

The relation §f(x)dx=F(x), written without the arbitrary 
constant, is used for formulas, etc., and when so written means : 
"An integral of f(x)dx is F(x)." 

In the expression \f{x)dx, f(x)dx is called the integrand. 

There is no systematic theory of integration as there is of 
differentiation; finding an integral is more a matter of search 
and discovery than of computation, but there are principles and 
rules to aid in the search. Moreover, as any formula of differen- 
tiation can be read as a formula of integration, we can begin by 
10 



130 The Calculus. 

making the following table of fundamental integrals, in which 
u is any variable, and may be a function of the independent 
variable : 

(1) \cdu — cu. 

u n+1 

(2) $u n du 



n+1 

(3) $e u du = e u . 

(4) U«du=^-. 

v ' J log a 

(5) f-^-=log«. 

(6) \co& udu — sin u. 

( 7 ) j sin ft,dft = — cos u. 

(8) {sec 2 ftdft = tanft. 

( 9 ) j esc 2 udu— — cot ft. 

(10) j sec ft. tan udu — sec ft. 

(11) j esc ft cot udu — — esc w . 

(12) jcot m?t£=log sin u— —log esc ft. 

(13) {tan udu — \og sec ft — —log cos u. 

(14) {esc t«Ztt= log tan-— = — log cot — 

(lo) —-=== = sin -1 1/, or = — cos -1 ft. 

v V 1 — ft 

(16) [-^-r=tan- 1 Mor = -cot- 1 w. 

v JH-ft 2 



(18) j" 



sec -1 W or = — esc -1 ft. 
-1 

6?ft 



V 2ft, -ft 2 

(19) {(/ 1 (ft,)+/ 2 (ft)+/ 3 (^)+ )du 

= $fi(u)du+$f 2 (u)du+$f 3 (u)du+ . 

(20) \udv — uv—\vdu. 



Integration. 131 

Any formula of integration (for instance, any of the identi- 
ties above) can be verified by differentiation, with the aid of the 
definition of an integral : 

d$f(x)- dx = f(x)dx. 

Exercise. — Verify each of the twenty fundamental formulas. 

115. Direct Integration. — It requires experience and careful 
observation to recognize these fundamental forms in all cases. 

\{a 2 -x 2 )U{a 2 -x 2 ) 
is an obvious instance of 

§u n du; 
but the precisely equivalent expression 

-2$x\ / a 2 -x 2 -dx, 
or the expression 

\x^\J a? — x 2 - dx, 

which is — \ as much, is not so obvious. 

The real difficulty is that it is not enough to be able to say that 
d(a 2 — x 2 ) is —2xdx; the mere presence together of (a 2 — x 2 ) 
and x must suggest this fact. 

As another instance of the same formula, consider 

J sin 6 cos d6. 

A mere reminder that d sin = cos dO is sufficient to make it 
evident that 

jsin 6 cos d0= jsin 6 d(sm 6) = i^31 . 

These will serve to point out the fact that, to practice inte- 
gration successfully, the student must be able to remind himself 
of the differential formula that will be useful. 

In order to keep track of constant factors, it sometimes is 



132 The Calculus. 

worth while to abbreviate the integrand by introducing a new 
variable. 

(x 2 
6 dx. 

We notice that x 2 dx is a constant multiple of d(x 3 ), and that 
x G = (x 3 ) 2 . Then let x 3 — y, so that 3x 2 dx = dy, and we have 

[x 2 d* _ [ jdy = 1 [ d(y/2) _ x , , y_ 
Jx 6 + 4 — J v 2 + 4 6 Jl + 0//2) 2 ~ 6 2 



V 

x 3 

X 3 

Substituting 1/=-^- would have made the work still more 
mechanical. 

116. Trigonometric Functions. — The fundamental formulas 
of trigonometry must also be so well known that useful relations 
will readily come to mind. For instance, the three important 
integrals 

\sm 2 6d0, $cos 2 6 dO, and j sin 6 cos 6 dO 

can be evaluated directly by means of the relations 

sin 2 = ±(l-cos20) ? 
cos 2 0=J(l + cos20), 
sin 0cos 6 = ^ sin 26, 
and 

fcos 26 d6=i$eos 26 d(26) =i sin 26. 
Again 

jsin 3 6 d6= - f (1-cos 2 0)d(cos 6), 
jtan 6 sec 2 6 dtf^Jtan 6 d (tan 6) = J sec 6 d(sec 6). 

117. Examples. 
Evaluate the following: 

1. \Vxdx. Ans. fVP. 

2. ]gtdt. Ans. igt 2 . 



Integration. 133 



dx A 1 



3 -^ _ 

5. \(x 2 -3x + 4:)dx. Ans. ^(2x i -9x 2 + 24x). 

!^_dx 
' )(x + 

, f ^ 

' J V3z + 



r4V • An «. t^tt 

1)1 (x+±)l 



Ans. |V3a; + 2. 



a:^ A — 1 

Ans. 



(a 2 + x 2 )¥ ~~" 3(a 2 +x 2 )^* 

9. ^W-z 2 dz. Ans. -i(a 2 -x 2 )'. 

10. $x 2 (a 3 + x 3 )*dx. Ans. i(a 3 + ar)i 

ll.J^. Ans. Iog(3+*). 



12. 



a^a; . , 1 

Ans. log 



a 2 -x 2 ' " ' " 6 Va 2 ^ 



x 

nx 



13. (a nx dx. Ans. -£ 

71 log a 

14. jze* 2 d:r. Ans. Je* 2 . 
6 d0= Jtan • d0. Ans. 2 log sec {. 



-I 1 



sin 



16. \ — — t, <^0. Ans. — log cos 0=k>2r sec 0. 

J COS 60 

17. fsin 2 0d0. Ans. |-J sin 20=J(0-sin cos (9). 

18. jcos 2 0d0. Ans. |+J sin 20=-|(0+sin cos 0). 

19. jsin cos d0. Ans. J sin 2 6, -J cos 2 0, or -J cos 20. 

20. jsec 2 0tan0d0. Ans. -J tan 2 or 1 sec 2 0. 

21. jsin 3 0<20. Ans. co ^ -cos0. 

22. J cos 3 J0. Ans. sin 0- ^—^ . 

23. jtan 3 0d0. Ans. ^-? -log sec 0. 



134 The Calculus. 

o< f dx f dx . . ,x—2 

M -)^5 + ± x - x z = )V 9-(x-2y Ans - sm ~T' 

25 - [ ,,f lK = L , /*, ,v 2 Ans. itan- 1 ^. 

J.T 2 + 2a; + 5 J4+(x + l) 2 2 

f cfo 

^•Jo-^Vz'-to+s- Ans - -s^ 3 -*)- 



28 
29 



J (1 + x) vx 2 + 2x 

30. f-^^ dO. Ans. -csc0. 
J sm 2 

31. f . f * ,. . Ans. log tan 0. 
J sm cos 

32. JVl-cos0<Z0. Ans. -2V2 cos |= -2V 1 + cos 0. 

33. jVl + cos(9J(9. Ans. 2Vl-cos 0. 

34. (sec ^^33 [ sec2 e + 8ec ei ^ e . ci6. Ans. log ( sec 6+ tan 6). 
J J sec + tan 

35. j (sec 0+tan0) n sec J0. Ans. — (sec + tan 6) n . 

118. Integration by Substitution. — An integral that bears no 
evident resemblance to any one of the fundamental forms can 
often be made recognizable by the introduction of a new variable. 
For instance, if in 



$Va 2 -x 2 dx, 
we put #=asin0, we have 



Va 2 — x 2 =a cos 0, dx—a cos 6 dO, 
and the integral becomes 

a 2 J cos 2 d6= ^- [0+sin cos 6] . 
(Ex. 18, Art. 117.) 



Integration. 135 

Hence 



jV 8 -A4[ S in-i + ^l-i 



or 



\ Va 2 -x 2 dx=i[a 2 sin" 1 — + xVa 2 -x 2 ]. 

The integral of any expression irrational merely through the 
presence of \/a 2 — x 2 , Va 2 + x 2 , or Vx 2 — a 2 can be rationalized 
by an appropriate trigonometric substitution : x — a sin 0, 
x=at&nd, or x = asecO. The resulting trigonometric integral 
is often recognizable. 

The integral of an expression irrational merely through the 
presence of V&x+b is generally recognizable, but is made 



2v 



simpler by putting vax+b = y, dx = — - dy. 

a 

119. Rational Fractions. — The integral of any rational frac- 
tion is made simpler by separating the fraction into partial 
fractions. (See Algebra, Art. 136.) 

For instance, 

dx 1 



}x 2 (x 2 + l)~ } x 2 . 

120. Examples. 

__ adx 

xA/a 2 -x 2 



= ■ — tan -1 x 



l+x 2 x 



„ f adx . , a—Va 2 — x 
1. 1 — / o . ■ Ans. log 



2. k d 2 X . Ans. \og(x+Va 2 + x 2 ). 

}ya 2 + x 2 

f fa 

3 - ]Vx 2 -a 2 ' AnS ' lo g(^+V^ 2 -a 2 ). 

4 - k /o ■ 2 = « Ans. log(x + a+ V2a.T+o^). 



136 



The Calculus. 



5. $x 2 Va 2 — x 2 dx. 

Ans. 



6. §xVa + xdx. 

7. [tt=^. 

J va + x 



8. 

9 

10. 



"a+x 
a — x 

dx 



dx. 



[ da 



l dx 



2 -l) 



" shr 1 — -~Va 2 -x 2 (a 2 -2x 2 ). 
8 a 8 v ' 

Ans. %(a+xyt-^-(a + x)L 
Ans. f(a+z)*— 2a(a + x)K 



Ans. — #— 2alog(a— x) 



A 1 t a+x 

Ans. >s - log 

2a G a— x 



Ans - '^ffi-^T) 



121. Areas Found by Integration. — Let it be required to find 

the area bounded by the parabola 
"B y 2 = 4x, the axis of x, and the line 
x = ±; the area OAB of Fig. 45. 

We shall proceed as follows: 
Drawing an ordinate, PM, through 
any point, P(x, y), of the parabola, 
we cut off an area, OPM, the extent 
of which is determined by the value 
A of OM—x, and varies when we vary 
x. This area, OPM, is therefore a 
function of x, as yet unknown. Call 
HF(x). 

Area OPM =F(x). 

We shall determine F(x) ; then the 

value of F(x) for any value of x 

will give the area cut off by the 

Fig. 45. ordinate corresponding to that value 




Integration. 137 

of x. In particular, when x=Q, the area shrinks to nothing; 
and when x=4, the area becomes the required area OAB. 

F(Q) = and F(4) = required area OAB. 

To determine F(x), we first find its derivative according to 
the general definition of a derivative. 

Extend OM to W , thus giving to x the increment MM'=Ax, 
and draw the corresponding ordinate P'M' ' = y + Ay. 

Then 

F(x) =area OPM, F(x + Ax) =area OP'M'. 
AF(x) =area MPP'M'. 

Complete the rectangles PM' and P'lf. 
Evidently 

PM'<AF(x) <P'M. 

PM' AF(x) P'M 

Ax Ax Ax 



or 



y<^<(y+*y) 



Now the desired derivative 

iF(x)l 

y> 



dF(x) _ \ AF(x) 
dx Ax 



AX=0 

for as Ax approaches zero, (y + Ay) approaches y, and the value 

of ^ is always between (y + Ay) and y. 

Finally 

dF(x) —ydx; 

and as y 2 = 4:X, 

dF(x)=2Vxdx, 
and 

F(x)=$2V^dx=%x?+C. 



138 



The Calculus. 



We have not yet entirely determined F(x), nor could we 
expect to do so by merely determining its derivative; for the 
process of finding the derivative would have been precisely the 
same if F(x) had represented the part of OAB between any 
fixed ordinate and AB. The ordinate at which F(x) becomes 
zero will, however, complete the determination of F(x) by de- 
termining the arbitrary constant C. For since F(0)=0 

O = F(0)=$> O + C; 
and C=0, so that 

Finally : 

The required area 0AB = F(4:) =£(4)C 3 ^. 

122. If it had been required to find the area bounded by 
y 2 = 4x, y = 0, x=l, and #=4, the discussion would have been 
just the same, except that the variable area F(x) would have 
been zero when x=l; i. e., we should have had F(l) =0. Con- 
sequently, although we should have had 

F(x)=fr£+0 

as before, the value of C would have been different, since 

0=F(l)=i+C 

gives 

e=-f. 

Then 

F(x)=i{xi-1) 

would have been the general expression for the variable area, 
and 

F(4) =|(8-1) =¥ 
would have been the required area. 



Integration. 



139 




a A M M 

ir 

Fig. 46. 



123. Consider now the general process exemplified in the 
preceding article. Let it be required 
to find the area AKLB, Fig. 46, 
bounded by any given curve, y = f(x), 
the axis of x, and the ordinates cor- 
responding to the abscissas a and h. 
Consider, first, the area AKPM 
bounded similarly, with the ordinate 
corresponding to the variable abscissa 
x in place of the one corresponding to 
the abscissa b. Denote this area, 
which is a function of x, by F(x). Then 

F(a)=0, F(b)= area AKLB. 
Increase x by MM' = Ax; y is correspondingly increased by 
Ay=PM'-PM, 
and F(x) by 

AF(z)=area MPP'M'. 

Complete the rectangles PM' and P'M. 
Evidently 



Divide by Ax : 



PM'<AF(x) <FM. 
AF(x) 



y< 



Ax 



<y+Ay. 



Therefore, 



AF(x) 
Ax 



A£C=0 



=y=f(x)> 



MM=f(x), d(F(x))=f(x)dx, 

[dF(x) is called the element of integration.] 
F{x)=Sf{x)dx+C; 



140 The Calculus. 

i. e., F(x) is any one of the integrals of f(x)dx plus some 
constant. 

Since.F(a)=0, 

0=F(a) = Uf(x)dJ\ mma +0, 

C=-[tf(x)dx] x=a , 
F(x) = Uf(*)dx-]-Uf(x)dx] x=a ; 

and the expression for the required area AKLB is 

Area = F(b) = Uf(x)dx] x=b -Uf(x)dx] x=a . 

In this expression for the area $f(x)dx may be any one of 
the integrals of f(x)dx, but must of course be the same one in 
both brackets. Evidently, changing this integral by a constant 
will make compensatory changes in the two brackets. 

For convenience, the expression for the area is more briefly 
written 

F{b) = \_Sf{x)dx}%% 

or, most conveniently, 



ATea = F(b) = \ b f(x)dx. 



The last form is read : " The definite integral from a to b of 
f{x)dx" In distinction, \f(x)dx is called an indefinite integral 
of f{x)dx. 

124. Definite Integrals. — Definite integrals have many other 
uses besides the determination of areas ; a general definition will 
therefore be useful for future reference. 

Definition.— If dF(x) =f(x)dx (i. e., if F(x) is an indefinite 
integral of f(x)dx), and if F(a)=0, then any other value of 
F(x) is 

F(b) = W(x)dx] x=h -Uf(x)dx-] x=a =\ b J(x)dx, 
the definite integral from a to b of f(x)dx. 



Integration - . 141 

An indefinite integral, \f{x)dx, is a function of x; the 

definite integral, f(x)dx, if a and b are given values, is the 
)a 

difference between two particular values of this function, and 

Cb 

so is a constant. If a and b are supposed to vary, f(x)dx 

will vary correspondingly : for instance, in the problem of the 
preceding article, any change in the abscissas a and b will cause 

the area f(x)dx to vary. In other words: 

Ja 

A definite integral is a function* of its limits. 

Besides the values of the limits, nothing affects the value of a 

definite integral, f(x)dx, except the form of f(x). For it 
Ja 

clearly makes no difference whether x is written throughout the 

integrand f{x)dx, or some other letter, since this letter, whatever 

it is, will be replaced by the limits when the integral is evaluated. 

For instance, each of the definite integrals, 



f a dx , f«_ dz 

Jo Va 2 — x 2 Jo Va 2 — . 



is (sin -1 1 — sin -1 0) or £. 

It is implied in the definition that the limits b and a written 
at top and bottom of the integral sign in a definite integral are 
values of the variable whose differential occurs in the integrand. 
This is important when a change of variable is made in evaluat- 
ing the integral. For instance, suppose we are to find the area 
of the circle x 2 + y 2 = a 2 or y — Va 2 — x 2 in the first quadrant of 
the coordinate axes. From the preceding article, this is 

A = j Va 2 -x 2 dx. 

In order to integrate, let x = a sin 6 ; then 



dx — a cos# d6 ; Va 2 -r — a cos 



142 


The Calculus 


when x=0, 






6>=sin- 1 = ( 


when x—a } 






0=siir 1 — — -77- 
a 2 



then 



H» 



cos 2 OdO = 



6 + sm0cos6 



•na 1 



(It is convenient to write the values at the upper and lower 
limits in the positions of those limits, and then subtract.) 

It is more expeditious to change the limits in this way when 
the variable is changed than to express A as a function of x and 
then substitute the original limits. The necessity of remember- 
ing the algebraic integral is also avoided. 

The notation of the definite integral can be used, to express 
any particular one of the indefinite integrals of a f(x)- dx; thus 

f(x)dx 

Ja 

represents the indefinite integral of f(x)dx that becomes zero 
when x — a. 

Thus we have seen that jsin cos d$ may be written 

jsin#d(sin0), -Jcos d(cos 6), or Jjsin 20 d(20), 

and is therefore 

sin 2 



COS" 



2 



or — 



cos 20 



These three values differ by constants, and of course there are 
any number of others. The particular one that becomes zero 
when 0=0 is 



sin cos 0<Z0=*HL 

2 



sin 2 fl 

2 



Integration. 143 



or 



or 



cos 2 

2 


cos 2 , j sin 2 

= 9 Tya 
* " 


cos 20 


] 6 cos 20 , 1 sin 2 6 
Jo" - 4 +*" 2 



125. Examples. 

Evaluate the following definite integrals: 



("2 
cos 3 a; sin zdz. Ans. J. 

o 



cos 3 2*da\ Ans. t. 

Jo 3 

ft __dx__ 
Jo V(a 2 -a; 2 ) 

f a x 

' Jo V(a 

JO COi 

f 1 !- 3 ^ * 
—F==dx. 

Jo Vl — a; 



Ans. -g-. 

4. . / , o 7- • Ans. a. 

x-) 



dO. Ans. i 



Ans. -2. 

7. Trace the curve a 2 y = ax 2 — x 3 and find the area of the 

a 2 
segment cut off by the a>axis. Ans. — — . 

8. Find the area enclosed by the curve, the z-axis and one of 
its asymptotes, given y 2 ( a 2 — x 2 ) = a 2 x 2 . Ans. a 2 . 

9. Find the area enclosed by the first arch of the curve 
y = smx and the ar-axis. Ans. 2. 

10. Find the area enclosed by the curve y=e x , the ?/-axis and 
the left half of the #-axis, and also the area bounded by the 
curve, the axes, and x—1. Ans. 1 and e — 1. 



144 The Calculus. 



11. V2ax-x 2 dx. Ans. ^ 

Jo i 

'/. v (2ax — x 2 )xdx. 

dx 
V{Qx~^x 2 -5) 

u -\lvWT¥) Ans - lo s^2+i). 



V (2ax — x 2 )xdx. Ans. -^- 

13. J" . ,,„ dx n „, Ans. r. 



a 
lb ' 



. f 2 V(a 2 -x*)dx. Ans. -|i (2tt + 3V3). 



126. Areas. — The result of Art. 123 may be stated : 
The area generated by an ordinate of the curve y=f(x) in 
moving from the line x~ato the line x—b is 



ydx—\ f(x)dx. 



It follows from a proof precisely similar to that of Art. 123 
that: 

The area generated by an abscissa of the curve x = f(y) in 
moving from the line y—a to the line y — b is 

f xdy=\ f(y)dy. 
Jo )a 

Thus the area bounded by y 2 — ^x, ?/ = 4, and y = 0, is 
I* 4 a f 4 v 2 j v 3 1 4 16 

Since the axes and the lines x — 4: and y = 4 bound an area of 
16 units, the area computed in Art. 121 is again seen to be 



(»- ¥)■ 



i«-¥Vf 



In practice, the required area can generally be found by tak- 
ing the element of integration parallel to either axis ; the choice 



2 




2 


a 2 


[■, ■ j. .1 


a 2 [0-0] 


2 


<f> — sin cos <£ 


-" " 2 Ll-oJ 



Integration. 145 

is determined by the relative difficulty of evaluation of j^a* and 

When a curve has convenient parametric equations, it is always 
best to use them, as a change of variable is thereby avoided. 
For instance, the area of the circle x 2 + y 2 = a 2 in the first quad- 
rant of the coordinate axes might have been found from the 
parametric equations 

x = acos<f>, y = asm<j>, 

where dx= —a sin <f> d<f>. 

The generating ordinate moves across this quadrant from left 
to right as 4> varies from \ to ; hence 

(a fO fO 

ydx—\ asm<f>( — asm<f>dcf))= — a 2 \ sin 2 <f> d<j> 

4 ' 

2 U * J 

127. Sign of the Definite Integral.— If $f(x)dx=F(x), 
P f(x)dx=F(b) -F(a) and f a f{x)dx=F(a) -F(b). 

Hence : 

Reversing the order of the limits in a definite integral changes 
the sign of the result. 

For instance, — a 2 \ sin 2 <f> d<j> in the preceding article is the 

k 

same as a 2 sin 2 <f> d<j>. 
Jo 

In deducing the expression f(x)dx=\ ydx for the area, 

it was tacitly assumed that f(x) or y was always positive, and 
that x increased from a to b. If f (x) is negative, the sign of the 
integral is reversed; and if x decreases from a to b, dx is nega- 
tive, and again the sign of the integral is reversed. 
11 



146 The Calculus. 

In other words, if the generating ordinate is negative or 
moves in the negative direction, I f(x)dx=\ ydx has a nega- 

)b [a 

ydx or 1 ydx gives the positive area. Of 

course, if both of these happen at once, ydx remains positive. 

}a 

To avoid trouble in practice, it is advisable in any problem in 
which such questions may arise to divide the area into separate 
parts so that y shall have one sign throughout each part, and 
then take the sum of the positive values of the separate areas. 

Of course, all these remarks apply to \xdy as well. 

128. Examples. 

1. Find the total area of the ellipse in four ways, using the 
single equation and the two parametric equations, and in each 
case taking the element parallel to the a>axis and also parallel 
to the ?/-axis. Ans. -rrab. 

2. Trace the curve x(a 2 + y 2 )—a z , and find the whole area 
between the curve and its asymptote. Ans. -rra 2 . 

3. Trace the curve x 2 y 2 (x 2 — a 2 ) =a 6 , and find the area between 
the curve and the asymptote, x — a. Ans. tto 2 . 

4. Find the whole area between the curve y(a 2 -\-x 2 )—a z and 
its asymptote. Ans. na 2 . 

5. What is the whole area between the curve x(a 2 + y 2 ) 2 = a? 
and its asymptote? 

Ans. — . 

6. Find the area between the witch xy 2 — 2ay 2 + 4=a 2 x=0 and 
its asymptote. Ans. 47m 2 . 

7. Find the area between one branch of the cycloid, x — 
a(cj> — sm<f>), y = a(l — cos<£), and the z-axis. Ans. Zira 2 . 

8. Find the area included between the cissoid x 3 -2ay 2 + xy 2 
= and its asymptote. - Ans, 3ira 2 . 



Integration. 



147 



129. Volumes of Revolution. — The area bounded by a curve 
y=f(x), two ordinates x — a and x—b, and the axis of x is 
revolved about the axis of x, generating a surface. It is required 
to find the volume enclosed. 

Let P(x, y) be any point of the curve; then the volume gene- 
rated by the area bounded by the curve, the axis, the ordinate 
x — a and the ordinate of P is a function of x, the abscissa of P. 
Call this volume F(x). To obtain its derivative, increase x by 
Ax; F(x) is increased by the element of integration AF{x), the 




Fig. 47. 



volume generated by the area bounded by the curve, the axis, 
and the ordinates of P and P'{x-\-Ax, y + Ay). P and P' 
generate circles whose radii are y and (y+Ay) ; the c}iinder 
having the smaller circle as base and Ax as altitude is smaller 
than the volume AF(x), and the cylinder having the larger 
circle as base and the same altitude Ax is larger than AF(x) ; 
hence 

7ry 2 Ax<AF(x) <Tr(y + Ay) 2 Ax, 



^ 2< ^? <7r( ^ +A ^ 2 ; 



so that 



dF(x) 

dx 



AF(x) 
Ax 



Ja.t=0 



: *2T 



148 The Calculus. 

or 

dF(x) —infdx. 

Since F(a) =0, the required volume is therefore 

V=F(b)=[ b 7rfdx. 

The revolution about the axis of x of the area bounded by a 
curve y=f(x), two ordinates x — a and x—b, and ihe axis of x 
generates a solid of revolution having the volume 



V: 



TT 



J y 2 dx. 



In the same way, it can be proved that : 

The revolution about the axis of y of the area bounded by a 
curve x = f{y), two abscissas y = a and y—b, and the axis of y 
generates a solid of revolution having the volume 



V=Tri b x 2 dy. 



130. Examples. 

1. Find the volume of the cone formed by revolving the line 
y— -j-~ x about the z-axis, the altitude of the cone being h. 

Ans. -3-. 

2. Find the volume of the sphere formed by revolving the 
circle x 2 + y 2 = a 2 about the z-axis,, and also by revolving about 
the 2/-axis. Ans. |-7ra 3 . 

3. Find the volume required in example 2, using the para- 
metric equations of the circle. 

4. Find the volume of the ellipsoid formed by revolving the 

ellipse — 2 -f -£2- = 1, (a) about the z-axis, (b) about the y-axis. 
a 

Ans. f 7r& 2 a; |-7ra 2 &. 



Integration. 149 

5. Find the volume formed by revolving about the a>axis the 
parabolic segment having the double ordinate 2b for its base 

and altitude a. Ans. — «-■ 

6. Find the volume of the hour-glass-shaped figure formed by 
revolving about the i/-axis the area enclosed between the para- 
bola of example 5 and the lines y — b and y——b. 

Ans. *£*. 
5 

7. Find the volume formed by revolving the witch 

ifx + a 2 x — a z — 

-2 a 3 

about its asymptote. Ans. 9 . 

8. Find the volume produced by revolving about the a:-axis 
the segment of the cissoid y 2 (2a — x)=x 3 cut off by x — a. 

Ans. 8a 3 7r(log 2-f ) = 0.6655a 3 . 

9. Find the volume formed by revolving about the z-axis the 
part of the curve y=e x lying to the left of the origin. 

Ans. f. 

10. Find the volume of a capstan 2b in height, the curved 
surface of which is formed by the revolution about the y-axis 

X 11 

of the hyperbola — Y — -p =1. Ans. f 7ra 2 b. 

131. Further Methods of Integration: Integration by Parts. — 

The formula $udv = uv — $vdu (20, Art. 11-1) is of great service 
in the integration of transcendental functions. 

For instance, required \x 2 log xdx: 

Let \ogx—u, and x 2 dx = dv; then 

7 (J/X C o 7 X 

du — - , v=\x' ! dx= -s- : 
x } 3 

therefore, 

j[log#] \_x 2 dx~\ — \udv — uv — \vdu 



150 The Calculus. 

Again, required jsin -1 x - dx: 
Let sin -1 x — u, and dx — dv; then 

du ~ T7T=% > and v ~ x ' 
Vl — x 2 

therefore 

xdx 



jsin -1 x ' dx^xsur 1 x 



Vl-x 2 
= x sin- 1 Z + -JJ {l-x 2 )-U(l-x 2 ) 

— x sin -1 x + VI — x 2 . 

132. Examples. 

Deduce by means of this formula the values of : 

1. jsin 2 dO, jcos 2 d$, jsin cos dO. 

2. ]x sin xdx=sinx — # cos x. 

3 . | tan -1 xdx = x tan -1 x — log Vl + x 2 . 

4. jxsin -1 xdx=-y- sin -1 # — J (sin -1 a;-jVl-a; 2 ). 

5. (x n logxdx= -flog a; — - r J. 

6. jcot cos d0=cos + log tan J. 

133. Trigonometric Functions. j sin /w0 cos n6 dO, etc. — 

These integrals are important in problems of Mathematical 
Physics, j sin mO cos nO dO is readily evaluated if we note that 
sin(ra + n)0 + sin(m.— n)0 = 2 sin mO cos nO. For instance, to 
find I— jsin 30 cos 20 dO. Since 

sin(30 + 20) =sin 30 cos 20 + cos 30 sin 20 
and 

sin(30-20) =sin 30 cos 20-cos 30 sin 20, 

the sum gives 

sin 50 + sin = 2 sin 30 cos 20. 
Thus 

7 = i|sin 50 d6 + lS sin d0= - T Vcos 50— J cos 0. 



Integration. 151 

jsin mO sin n6 d$ and J cos mO cos nO dd are treated in the same 
way. 

134. Integrals Containing Powers of the Trigonometric Func- 
tions. — We have in Art. 114 the integrals of the first powers of 
all the trigonometric functions, for 

jcsc tfflfclogtan | —log (esc 6 — cotB), 

and, since csc(£ + 0) = sec and d(l + 0)=d0, 

|sec 0d6-log tan(\ + f ) = log (sec + tan 0). 

"We can integrate any positive odd power of sine or cosine, 
any positive integral power of tangent or cotangent, and any 
positive even power of secant or cosecant by using the formula? 
connecting the squares of the trigonometric functions : 

sin 2 + cos 2 = 1, sec 2 — tan 2 = 1, esc 2 — cot 2 = 1. 

In the following discussion, n represents a positive whole num- 
ber, so that 2;i + l is an odd number, 2n an even number. 

jsin 2 " +1 • d$= jsin 2 " sin d0 = J (1 -cos 2 0) n sin dO, 

which can now be expanded and integrated. 
In the same way, 

jcos 2 " +1 J0= j(l-sin 2 6) n cos dO. 
jtan" d0= jtan"" 2 0(sec 2 0-1) d0 

= jtan"- 2 d(tan 0) - jtan"" 2 dO 

11 — 1 
Kedueing jtan 71-2 d6 in the same way, and so keeping on, 
until we reach either 

jtan 0d0=$d0 = or J tan d0 = log sec 0, 
the integral is completed. In the same way, 

Jcot" 0d6=- jcot"- 2 d(cot 0) - jcot"- 2 d0 

= - cotn ~\ e -Scot?-* dO. 

n — 1 



152 The Calculus. 

J sec 2 * dO= J (1 + tan 2 0)*- 1 sec 2 dO. 

(1 + tan 2 0) n_1 may be expanded and the powers of tan in- 
tegrated by the method just described. 

135. We can integrate any even power of sine or cosine by 
rising the formulas 

sin 2 0=i(l-cos20); cos 2 0=J(l + cos 20). 

$cos 2n 6 d6=~ j(l + cos20) w d0=-^ r j(l + cos<£)*d<k 

where <f> = 26. The even powers of cos <f>, got by expanding 
(1 + cos cf>) n , may be treated again in the same way, and the odd 
powers may be treated by the method above. 

Jsin 2 »0d0=-L j(l-cos <£)«#, 

where <j> — 26, is similar. 

136. Examples. 

ZL — 

1. [ ¥ tan 3 0d0=:i-Jlog2. 2. f * sec 4 d9=$. 

Jo Jo 

rr n_ 

3. sec 4 a;tana:^=- 1 4 5 -. 4. tan 5 xdx — \ + log 2. 

Jo Jo 

5. f 2 sin 2 cos 3 dO = ■&. 0. f 2 sin 4 0cos 3 «W=A- 

77 7T 

7. f 4 tan 2 xsec 4 ^a: = T 8 5. 8. ( * tan 4 d6-\ — \. 

9. Jsin20sin0d0 = ^sin0-isin30. 
10. \ cos 20 cos 40 = \ sin 2(9 + T V sin 60. 



n.j 



cos* 6 d6= g^ 



12. Find the area of the segment of the curve x 2 (a 2 — y 2 ) =a* 
cut off by x = aV2. Ans. 2a 2 (l-f ). 



Integration. 



13. Find the area enclosed by y(a 2 — x 2 )—a z ,y=Q f x——^ 7 
and x— — . Ans. a 2 log 3 = 1.1a 2 , about. 

137. The only powers of trigonometric functions left are odd 
powers of secant and cosecant. These can be integrated by using 
the relations sec 2 — tan 2 = 1 (or esc 2 — cot 2 = 1) and integration 
by parts. The integration of sec 3 follows : 

I={sec 3 0={sec0-sec 2 0d0, 

and in \udv — uv — \vdu, if 

u = sec 6, dv = sec 2 dS, 
we have 

du = see 0tan 6 dO, v = taoo.$, 
and 

7 = sec tan 0- {sec tan 2 d$; 
and since tan 2 — sec 2 6 — 1, 

7 = sec tan 0- {sec 3 d0+ {sec dO 
= sec tan 0-7 + {sec 6 d6; 
so 

27 = sec 6 tan 6+ {sec d6=see 6 tan + log (sec (9 + tan 0), 
7= {sec 3 d0 = -£[sec tan 0+log(sec + tan 0) ]. 

138. Examples. 
Prove in this way : 

1. {esc 3 d6 = ±[-csc cot + log(csc 0-cot 0)]. 

2. jsec tan 2 0~tf0 = i[sec tan 0-log(sec + tan 0) ]. 

3. jese cot 2 d0=^[-csc cot 0-log(csc 0-cot 0)]. 

4. Prove each of the last four formulas by beginning with the 
relation of the squared functions and following with integration 
by parts. 

5. Show that [° \fa 2 + x 2 dx= ~ [V2 + log(l + V2)]. 

Jo ^ 



154 



The Calculus. 



6. Show that (sec 2n+1 6 d0= -£- sec 2 "" 1 6 • tan 
J 2n 



2n-\ 

2n 



jsec 2 "" 1 • dO. 



7. Prove Jsec 5 • c?5 = J sec 3 tan + § [sec (9 tan 

+ Iog(sec0 + tan0)]. 

8. Show that the segment out from y 2 = 2ax + x 2 by x = 2a has 
for its area a 2 (6\/2-log(3 + 2V2)). 

139. The method of Art. 137 furnishes a convenient formula 
for certain definite integrals, for instance : 



f 

Jo 



sin" dO, and | cos" dO, 
o 



where n is any positive integer greater than nnity. 



-I; 



I=\ sm n 6'd0=\ sin"" 1 • sin 6 • dO. 

Call u = sm n ~ 1 6, dv = sin dO; then 

du— (n — l)sin n - 2 0cos 8 d9; v=—cos6. 



1 = 



sin n_1 6 cos 6 



+ 



it 



l)sin*- 2 0cos 2 0d0. 



The bracket is zero, and cos 2 6=1 — sin 2 6. 

IT 

1 = 0+ (n-1) f 2 sin"- 2 0-tf0-(?i-l) 
j = ( n _l) f 2 sin"- 2 0-dO- (n-1) J; 
w ./=(n-l) ( 2 sin«- 2 0-rj0; 

£ — 

I=[ 2 zm n 0d6= — - [* 8in»- 2 0d0. 



sin n 0d0; 



n jo 









Integration. 




By 


this formula, 










■a 

I 


sin n - 2 


n 

6d6= r ^[ Y sii 
n-2)o 


i^OdO; 


SO 












It 

F • 

sm" 
Jo 


ede-. 


_ ( n -l)(n-3) 
n(n — 2) 


n 

sin"- 4 1 
lo 



155 



ode. 



If this process is kept up, the last step will finally be : 
If n is even, 



■2 fT 

If n is odd, 



sm$de=l. 

Jo 

Hence 



[ 2 sin 0d0=[ 2 d0=~ 
Jo Jo 2 

it 

c 



is even 



ir it ff 

f 2 cos n 0d6= j 2 sin"(-J-^)^=- f° sin n <j> d<f>= T sin n <f> dcf>, 



if f -0 = <£ and d6=-d<t>. 
Hence 



f 2 eos" </0 = f 2 sin" <£ </<£ = [ 2 sin" </0. 
Jo Jo Jo 

These results should be memorized; they are very often useful. 



156 



The Calculus. 



140. Examples. 

1. Evaluate the following definite integrals by the formulas of 
Art. 139. 

— !i 

(a) j* sin 4 = ff , (b) J* sin 5 d$= T \, 

IT 

(c) ( V sin 5 de= [ 2 sin 5 d6 + I 



sin 5 d$. 



(d) 



'2" 



77 

cos 5 Ode, (e) [* cos 5 d0 7 (f) [* 



cos 6 6?0. 



2. Find the volume generated by revolving the cycloid 

x = a(cf> — sin <£), y = a(l — cos <£) 

about its base. [Hint, in y and dx, change (1 — cos<£) to 
2 sin 2 *, and % to 0.] Ans. 57r 2 a 3 . 

3. Find the volume formed by the revolution of the curve 
xy 2 + 4cCL 2 x = 8a 3 about its asymptote. Ans. 4nr 2 a 3 . 

141. Similar formulas may be developed for the integral of 
the product of a power of sine by a power of cosine. 



i> 



sin m 6 cos n 6 dd-. 



n 

r 



sin™ cos"" 1 cos <Z0. 



Call sin™ cos dB-dv, cos"- 1 = w; then 
sin m+1 



— J: , du= — (>-l)cos"- 2 0-sin0 dd. 



_ f |" sin m+1 cos"" 1 
~ [ [ m + 1 



2 


-1 



m+1 . 
By the same formula, 



1 r 2 cin»» +2 a 
0l + ^£L_^( n _i) CO s«- 2 0.^0 

J Jo m + 1 

r 

sin» l+2 cos"" 2 d0. 



2 sin w+2 cos"" 2 <Z0 = - — 1 f 2 sin m+4 cos"" 4 dO. 
lo w + 3 Jo 



Integration 



157 



Continuing this process, we finally have: 
If n is odd, 



1= 



where 



(ra + 



7T 

1 ( w~ 1 il?~ 3) yi 2 5T f 2 sin "' + "" 1 6 cos de 5 

l)(m + 3) .... (m + n — 2) Jo 



71 

i; 



sm r 



6 cos 6 d8 = 



sin m+ " 



ra + ?t 



m + n ' 



and thus, 



T sin- e cos- * <w= / ( ?r/' )(n - 3) --;- 2 , 

lo (ra + 1) (ra + 3) (m + n] 

If n is even, 



1= sm m 0cos n 0d6 



^1)1^- 3) (n- 5^^1 f* rf cog0 

)(ra + 3)(ra + o) (m + w — l)Jo 



(.TO + 1) 

and two different cases arise, according as m is odd or even. 
If m is odd, (m + n) is also odd, and 

(n-l)(n-3) ....1 



7= 



(ra + 1) (ra + 3) (m + n-1) 



X 



(ra + K-1) (m + w— 3) 2 

(ra+ra) (ra+ /i — 2) 3 

If ra is even, (m + n) is also even, and 

j_ (w-l)(w-3)(n-5) ....1 

(ra + 1) (ra + 3) (ra + 5) (ra + ?i — 1) 

(ra + n — l)(ra + n — 3) 1 r 

X (ra + n)(ra + n-2) ....2 X 2 * 

In the last two formulas, (ra + 1) (ra + 3) .... (m + n — 1) is 
the product of the highest of the factors in 

(m + n-1) (m + n-?,) (1 or 2) ; 



158 The Calculus. 

hence the formulas are: 

IT 

[ 2 sin" l 0cos n dd = 

(n-l)(n-3)(n-o) 1- (m-1) (m-3)(m-5) 1 5 

(m + n)(m + n — 2)(m+w-4) 2 ' X 2' 

w;7jm m and n are both even; 
[^ sm m 0cos n 0d0 = 

(n— \){n— S)(n— 5) (2 or 1) • (m-1) (m— 3)(m-5) (2 or 1) 

(m + n)(m + n— 2)(m + w— 4) (2 or 1) 

w7iew m and w are no£ both even. 

These formulas should also be memorized; they contain the 

formulas for sin m 6 dO and cos n dO as special cases in 
Jo Jo 

which one of the two exponents in sin m cos" dd is zero. 

Jo 

142. Examples. 

1 . T sin 3 cos 4 e de= %^~ = A = r cos3 6 sin * * <**• 

Jo 7 • 5 • 3 35 Jo 



2. rsin 2 0ecs 5 0J0- 



7T 

sin 4 
Jo 



cos 6 6 d$: 



105 ' 
3r 



512 

- 
sm 3 0cos 5 0d0=-i-. 






5. I sin 2 0cos 2 0£70= T Y 



Integration - . 159 

6. Find the area of the curve ( — V + l-^j- j =1, using para- 
metric equations. (See Art. 87.) Ans. %Trab. 

7. Find by means of parametric equations the areas of the 
curves a 8 y 2 = x 4 (a 2 — x 2 ) 3 and a f< y 2 — x H (a 2 — x 2 ). 

Ans. -^- for each. 

o 

8. Find the area between ( — J -f- (-j-\ = 1 and the coordinate 

. ab 

axes. Ans. •—%- . 

b 

143. The indefinite integral of the product of two powers of 
sine and cosine can always be got directly, unless both powers are 
even. For 

jsin 2 '" +1 cos" 6 d0 = - J (1-cos 2 6) m cos" 6 d cos 6 
or 

jcos 2m+1 sin" 6 de= \ (1-sin 2 6) m sin n 6 d sin 0. 

When both powers are even the use of the double angle will 
always simplify the integral. For if m < n, 

I- jsin 2w cos 2n d9= \ (sin 2 " 1 cos 2 '" 0)cos 2 <"- w) 6 d6; 

since 

. n n sin 20 , Q l + cos?# 
sm 6 cos 0— - — - — , cos 2 8= , 

I= 22w+ (L,) + x lsin 2 -(2^)(l + cos2^)"-^(2^) 
= g^Li Jsin 2 '"<£(l + cos4>)"-'"^ 

where <£ = 20. 

A similar reduction gives, if m>n, 

I=$sm 2m 6cos 2n 0d8= — Ls j sin 2 n <£(1- cos <£)'""" <7<k 

where <j> = 20. 



lfi The Calculus. 

Whenever the powers are equal, it is well to note that 

1 = Jsin" cos" d0= -L j s i n n ^ d ^ 

where <f> = 26. 
For instance, 

I ± = J sin 2 cos 3 d0 = Jsin 2 0(1 -sin 2 0)d(sin 0) 

__. sin 3 _ sin 5 
3 5 ' 

I 2 = Jsin 2 cos 4 J0= J (sin 2 cos 2 0) (cos 2 0)670 
= ^j fsin 2 0(1 + cos <£)<?<£, 



or 



J 2 = -|f (20-sin20cos20)+ 2^7sin 3 (20). 
I 3 = Jsin 2 cos 2 d0= -~ j sin 2 <£ d<£ 

= i(20-sin20cos20). 

144. Examples. 

1. Show Jsin 3 0cos 3 <W=-i- (J cos 3 20-cos 20). 

2. Show jsin 4 cos 4 670=^- (60-2 sin 40+J sin 80) 

3. Find P cos 2 sin 3 dO. Ans. ¥ Vt5 

'77 



5 7T 



4. Find 



sin 4 cos 4 d0. Ans. ^g.. 



6" 

145. Quotients of powers of sine and cosine may be somewhat 
similarly handled, but are often best expressed in terms o1 
secants, tangents, cosecants, cotangents. 



Integration. 161 

For instance, taking advantage of the fact that 
d tan 9 = sec 2 9 dO, d cot 9 = - esc 2 9 dO, 

f^4 d0= (tan 3 9 sec 2 9 d$= (tan 3 d tan 9 = i ^~^ . 
J cos 5 9 4 

This method serves when the degree of the denominator is 

any even number greater than that of the numerator. 

Taking advantage of the fact that 

d sec 0=sec 9 tan 9 dO, d esc 9 — —esc 9 cot 9 d9, 

we can handle any quotient of this sort having a numerator of 
odd degree, and a denominator of higher degree. For instance. 

9 dO = jtan 2 9 sec 9 tan dO 



COS 4 9 



= S (sec 2 6-l)d sec = ^^ -sec0. 



f sm 3 ^ ^_ . tan2 ^ gec e tan ^ gec ^ de 

}cos 5 9 J 

r / 3 a n\ j n sec 4 9 sec 2 

= j (sec 3 9 — sec 0) a sec 9 — — — — — -— . 

These methods cover so far all cases where the denominator is 
the term of higher degree, except the one where the numerator 
is of even degree and the denominator of odd degree. These 
may be handled as follows : 

fsinl* d$= fsini| $= [jr^idx, 

J cos 5 9 Jcos 6 J(l — x) 

where x = sin9. This fraction may be broken into partial frac- 
tions by the ordinary algebraic methods and the parts integrated. 
Another way is to write the integral as 

jtan 2 9 sec 3 9 d0= $(sec~° 9 -sec 3 9)d9. 

As any power of secant can be integrated, this method is also 
invariably feasible. 
12 



162 The Calculus. 

146. There remain the cases in which the numerator is of 
higher degree. The following examples show methods which 
can always he made to work: 

J cos 2 6 J cos 2 9 J cos 2 9 

where a;=cos 9. 

\^e de= f (i-ooB'tf)' ^ =I ( ^ os2 )M 

J cos 2 9 J cos 2 9 J v y 

9 <Z0= jtan 3 sin d$= -cos tan 3 + J3 tan 2 sec d0 



[ sin 4 
Jcos 3 

(integrating by parts). 

147. Examples. 

L f 81 ^ d9 = $ (sec 3 9 -sec 9) d9 
J cos 3 J v y 

=J[sec tan (9— log (sec 0+tan 0)]. 

2 [ CQs3 x dx— csc3 ^ _ csc5 ^ 
' J sin 6 a: ~~ 3 5 

o [ dx f sin a^£ -, / , * 

<*• - — - = hi ^— = log(csca; — cota). 

J sin a; J 1 — cos 2 x & v ' 

4. M^ =Hsec0tan0 + log(sec0+tan0)]. 
, f ^ +/) cot 8 

6. f^> = f- cos ^ 1 r/f ,g ) a ^=i(cotgco8ec'g-2cot^+ 

J sin 3 J sin 3 

cos sin 0) d0. 

(d0 
= — are readily handled 
a + b cos 9 J 

by means of the functions of the half-angle; an example will 
show the method. 



Integration. 163 



2 = 2 cos 2 1 +2 sin 2 1 



j2 + 3cos0 - " *-r~ m ~~ * 

3 cos 0=3 cos 2 f -3 sin 2 1 

2 + 3 cos = 5 cos 2 f — sin 2 | 
Let | = </>; then 



f dO f 2^ 

j2 + 3eos0 ~ J5cos 2 (£-sin 2 (£ 

( 2 sec 2 <fr Jcfr _o[_^L 
5-tan 2 £ " Jo-a; 2 ' 



where £ = tan <6 = tan|. 



2 \/5+x _1 V5+tan| 

2V5 l0 S ^Z^ = V5 l0 S V5_tan| 



tf0 



Jfl — 6 cos </> ' 



|a+6 sin 
where <£ = £ + 0. 
The only very important integrals of this type are 

f dO f dO f J0 f J0 

Jl + cos0' Jl-cos0' Jl + sin^' Jl-sin^' 

all of which are directly integrable. 

149. Examples. 

Integrate the following: 

2 [ ^0 -il 2 + tan j 
* j3 + 5cos0 ~~ 4 g 2-tanf 

3. — =tan? = csc — cot 0. 

J1 + COS0 2 

4. f . de -- C Qtf=-CSC0-COt0. 
J 1 — COS 2 

5 * |l4ln^ = - cot (?+')= taI1 ^- sec ^ 



164 



The Calculus. 



dO 



l-sin0 



=:tan (?+?) =tan + sec 6. 



150. Representation of an Integral by an Area. — We have 
seen that the area bounded by a curve, y = f(x), the two ordi- 
nates x—a and x=b, and the z-axis is given by the definite 

integral A= P f(x)- dx. (See Arts. 124 and 127.) 

Conversely, this area may be used as a graphic representation 
of the definite integral. The integral whose value is thus rep- 
resented by an area may be itself an area,. a volume, or any other 
of the many sorts of quantities that are computed by integration. 

151. The Limit of a Certain Sort of Sum. — Quantities such 
as we have determined by integration can be found in another 
way. For instance, 

To Find the Area Bounded by /= - — , y=0, and x=a. — 

Divide the area into n strips by equidistant ordinates Ax= — 

apart. Call the distances from 
J\ the origin to the points of divi- 
sion in OX: 

x — 0, x x — Ax, x 2 — i lAx, 

. . . . , x n = nAx = a, 




MM' 



Fig. 48. 



and the corresponding ordinates 



2/o = = 



Vi = 



%i 



y 2 = 



y n = 



Inscribe a rectangle in each strip and circumscribe one about 
it; the sum of the inner rectangles is less than the required 
area ; the sum of the outer ones is greater ; and the two sums, 



y Ax+y 1 Ax + y 2 Ax + .... +y n . 1 Ax 



Integration. 165 

and 

y 1 Ax + y 2 Ax+ .... + y n Ax 
differ by 

(y n -y )Ax. 

As Ax approaches zero as a limit, this difference approaches the 
limit zero, and each of the sums therefore approaches the re- 
quired area. The first sum is 

n A , (Ax) 2 . , (%bx) 2 A , ["(rc-l)Aafl 2 A 

0«Aa;+A L- Ax+ ± '—Ax-\- .... + -^ * ±- Ax 

a a a 

= M![o + 12 + 2 2 + 3 2 + .... + (rc-l) 2 ] 
o 

= \nj_(n=l) (fltt _ 1)ln)=a ,\_nlU_ 1\ (1 



a \ 6 
The second sum is 



:){h-l){.)=^ ^_I (,). 



(A3;) 2 . (2Ao;) 2 A , (3Aa;) 2 . , , (nAo:) 2 A 

■^ £- Ax + -* — Ax + ± '— Ax+ .... + -* '— Ax 

a a a a 

_ (AxY_ (1 2 + 2 2 + 3 2 + +n2) 

= (il(l)^ +1 ) (re+ i),^)( 2+ i)( 1+ A). 

The common limit approached by the sums as Ax approaches the 
limit zero and n consequently increases indefinitely is 

a 2 (4)(2)(l)=-J, 

the required area. 

The same method can be used for volumes; indeed, except for 
the notation, this is the method already niade familiar in the 
study of Solid Geometry, where it was used to find the volume 
of the pyramid, cone, and hemisphere. Any quantity that can 



166 The Calculus. 

be determined by a definite integral can be regarded as the limit 
of a sum of this sort. The advantage of the new conception of such 
problems lies in the saving of much preliminary discussion. The 
determination of the limit by algebraic means would be laborious 
and in many cases impracticable; but it is also unnecessary, for, 
as we shall see in the next article, the actual computation can 
be made by evaluating a definite integral precisely as in the 
earlier method. 

152. The Definite Integral as the Limit of a Sum. — Any of 

the problems of which we have been speaking may be described 
as follows: A variable x is divided at the values x , x x , x 2 , x 3 , 
. . . . , x n , so that the quantity (x n — x ) is divided into n parts, 
namely, 

lA,/. q — JU-t Xq+ ZAt.t'-f — — Xn X-\y IjktlsQ — Xq X<£J • . . . , 

and either one of two sums is formed: 

f(x~ )Ax + f(x 1 )Ax 1 + f(x 2 )Ax 2 + .... +f(Xn- 1 )Ax n - i , (1) 

in which each term is the value of f(x) at one of the points of 
division multiplied by the following Ax, or 

f(x 1 )Ax + f(x 2 )Ax 1 + f(x g )Ax 2 + .... +f(x n )Ax n _ 1 ,\ (2) 

in which. each term is the value of f(x) at one of the points of 
division multiplied by the preceding Ax. 

The number, n, of parts is then supposed to increase in- 
definitely, the sum of all the parts remaining equal to (x n — x ) 
and the value of each part approaching zero as a limit. In any 
such case, each of the sums (1) and (2) will approach as its 
limit the value 

n f(x)- dx. 

To prove this, consider the graph of.f(x), and the area A 
bounded by the curve y = f(x), the #-axis, and the two ordinates 



Integration. 



167 



x=x and x — x n - (Fig. 49.) Divide the part of the z-axis from 

x to x n into n parts corresponding 
to the values Ax , Ax 1} . . . . , 
Ax n _ 1} and erect at the points of 
division the (n + 1) ordinates, of 
which the lengths are f(x ), f{x x ), 

- , f(x n ). The area A is thus 

p IG# 49. divided into n strips. 

Each term of the sum (1) is the 
area of a rectangle inscribed in one of these strips, and the cor- 
responding term of the sum (2) is the area of the rectangle 
circumscribed about the same strip. 
(Fig. 50.) The first sum is there- 
fore the area of the polygon in- 
scribed in A } and so is less than A; 
and the second sum is the area of 
the polygon circumscribed about A , 
and so is greater than A. 

The difference between the two 
sums is the area of the small rec- 
tangles. If the divisions along the #-axis are all equal, this 
difference is 

A*„[/(0 -/(*„)]• 

If the divisions are unequal, let the largest of them be Axi; 
then this difference is less than 

Ax i [f(x n )-f(x )]. 

In either case, as n is indefinitely increased, Ax or Axi ap- 
proaches zero as a limit, and the difference between the two sums 
therefore approaches zero as a limit. Consequently, as one of 
the sums is always greater than A, and the other always less, 
each of them approaches A as a limit. But the value of A is 
already known to be 




Fig. 50. 



168 The Calculus. 

A = \ f(x)dx. 

Hence the limit of either the sum (1) or the sum (2), asn 
is indefinitely increased, is 

" f(x)- dx. 

153. As an illustration of the use of the definite integral 
regarded as the limit of a sum, consider the following problem 
It is required to find the volume produced by revolving about 

the axis of x the part of the parabola -~ = — between x— — 




Fig. 51. 



and x=a. Divide the volume into n slices by planes perpendicu- 
lar to the axis, at the following distances from the origin : 



x , x 19 x 2 , 



. • , Xn-D Xn — d, 



and let 

X 1 Xq — &Xq } X 2 X-l — A3Tj , . . . . , Xn Xn-i = &Xn-i* 

The corresponding ordinates of the parabola are 



Integration. 169 

Inscribe in each slice a cylindrical disc; the sum of all these 
discs is 

Try 2 Ax + 7rtj 1 2 Ax 1 + .... +Try 2 n -.Ax n -\ 
or 

— x Ax -\ x 1 Ax 1 + .... +—-a: n _ 1 Aa; n _ 1 . 

ci a a 

The limit of this sum, as n increases without limit and each 
division of the #-axis decreases without limit, is 

rr^ rH a = Wnab 2 
]a/i""""~~ a 2 \ a/i ~ 32 

It is obvious that the limit of the sum of all the inscribed discs 
is the volume of revolution required; it is, however, logically 
conceivable that it may be something less than the required vol- 
ume. But if we consider in the same way the discs circumscribed 
about the n slices, we have for the sum of their volumes : 

b 2 t:6 2 -b 2 

7T — x ± Ax + — x 2 Ax t + .... H x n Ax n . 

u.1 a a 



f*« nb 2 , Ttb 2 [ a , 

— xax— — xax— 

ho & a }al 



The limit of this sum is also 



f*« 7zb 2 7 -b 2 ( a , 

— xax = — xax 
)x a a }a/i 

the required volume oi 

must be the required volume. 



lo-n-ab 2 



and is either the required volume or something greater; hence 
IhTrab 2 

32 

We shall not take the trouble in later applications to remove 
this logical doubt, as it never interferes with the clearness of the 
discussion, and can always be treated in the same way. Further- 
more, we shall abbreviate the discussion by speaking of only a 
typical term of the sum whose limit we take. The preceding 
proof, thus abbreviated, is as follows: 

Divide the volume into slices by planes perpendicular to the 
x-axis, Ax apart, and in each slice inscribe a cylindrical disc, the 
volume of which is 



170 The Calculus. 

ttv^Ax = xAx. 

v a 

The sum of all these discs is an approximation to the volume 
required, and its limit, 

— xax. 
Ja/i a 

is exactly the volume. Hence the volume is 



f« Ttb 2 

— a 

J a/4 a 



/[;= iw6: 



154. The typical term of the sum whose limit is the definite 
integral is precisely the element of integration of which we spoke 
in using the earlier method of finding areas and volumes. In 
this later method, the parts into which the approximate area or 
volume is divided, the values of which are the terms of the sum, 
are all called elements of integration. 

dx, dx , dx x , dx 2 , etc., are frequently written in place of Ax, 
Ax , Ax 19 etc., for the infinitesimal factor of the element of in- 
tegration. When all these are equal, and each is represented by 
dx t the typical term of the smaller sum is f(x)- dx and the cor- 
responding term of the larger sum is f(x + dx)- dx. It will be 
seen by what has just been done in the preceding example that 
in order to be sure that the proof of any such problem can be 
completed rigorously, it is merely necessary that the true value 
of any one of the parts into which we have divided the area or 
volume that we are computing shall be intermediate in value 
between the corresponding elements, f(x)dx and f(x + dx)dx. 

155. The conception of the definite integral as the limit of a 
sum simplifies the consideration of certain general principles of 
integration, notably the connection between the integrals: 

7T 77 7T 

[" sin" 6 d6, f 2 cos n 6 dO, P sin™ 6 cos n dO, 



Integration. 171 

and the integrals of the same functions when the limits are whole 
multiples of — . 

An example or two will show the method. Consider 
j V sin 3 6 cos 4 dO. 

It represents the limit of a sum. The function 

sin 3 6 cos 4 e 

passes through a certain set of values as 6 varies, the sign of 
each term of the sum depending on sin 3 6, since cos 4 6 is always 
positive. Each term, therefore, from — £ to 0, is negative, and 
the terms are repeated in reverse order with positive sign from 
to \ , the corresponding terms of the two quadrants thus cancel- 
ling each other. The quadrant \ to -n- remains, each term of 
which is positive, and equal to the corresponding term in the 
quadrant from to f; hence 



sin 3 cos 4 6 d0= T sin 3 6 cos 4 6 d0 = 



¥ _-_. - „"_ A - ^ . 2 -3-1 

•5-3 ~ ' 35 



2 

By the same sort of reasoning 

[* sin 3 6 cos 3 6d6 = 0, 

as in the second quadrant cos $ is negative ; but 

3tt w 



f 2 sin 4 $ cos 4 8dO = S F sin 4 6 cos 4 6 dO 



_ 3 3- 1-3- 1 - 9- 



8 • 6 • 4 • 2 2 256 ' 
since the even powers are positive even-where. It is necessary 
in such cases merely to observe the sign of sin m 6 cos n in each 
quadrant. Where this sign is negative, the integration through 
the quadrant gives 



172 The Calculus. 

77 

- f 2 sin™ cos n 6 dO; 
and where it is positive, 

77 

+ P sin™ cos" (9 d0; 

156. Examples. 

1. Draw, very roughly, the graphs of 

sin 3 x cos 4 x — y, from — \ to 7r; 
sin 3 a; cos 3 x — y, from to 77-; 

Q 

sin 2 a; cos 6 x — y, from to ^ . 
v> 2 

Note how the relations just discussed are exhibited by the 
graphical representation of the integrals. (See Art. 150.) 

Find the following integrals : 

77 

2. T sin 2 cos 3 6 d$ = T \. f* sin 2 6 cos 3 6 d$= 0. 

3. [ n sin 2 cos 4 Bde-^. 

4. T sin 3 0cos 2 0d0=O; Psin 3 (9 cos 2 d0= T \. 

3tT 77 

("2" f f*H-D "2" 

5. sin 3 0cos 3 0d0= T V; sin 3 0cos 3 d0= ± T V- 

J 77 J tt77 



157. Areas with Curvilinear Boundaries. — The following 
problems are merely an extension of the ordinary problem of 
finding areas. 

I. To Find the Area of the Ellipse 4/2-4*/+ 17*2 + 12/ 
— 86*4-73 = 0. — Solving the equation for y, we get 



y = i(x-S) ±2V -x 2 + 5x-4:. 



Integration. 



1T3 



The curve is sketched in Fig. 52. Divide the area into vertical 
strips Ax apart ; the height of 
the strip corresponding to 
am r value of x is the sum of 
the two equal distances of 
the two corresponding points 
of the curve from the diam- 
eter y—\{x — 3), or is 



2x2V-x 2 + 5x-4:. 

We might get at this as fol- 
lows: The area is bounded 
above by the graph of the single-valued function 




Fig. 52. 



y 1 = i(x-3)+2V-x 2 + ox-4:, 

and below by the graph of 

y^=i(x-S) -2\/-x 2 + 5x-4:. 

The height of a strip is the algebraic difference of corresponding 
ordinates of the two graphs, or 

y t — y 2 = 4 V^-^-J-5^—4. 

The limits of integration with respect to x are the least and 
greatest values of x corresponding to points within the ellipse; 
as V— x 2 + 5x — 4 = y (x— 1) (4 — x), these values are 1 and 4. 

The area of the strip corresponding to any value of x is 

4V— x 2 + 5x — 4 • Ax, approximately, and the required area is 

exactly J 

Ci 

A- 4cV-x 2 + 5x-4:'dx. 
.1 

Since 



y-z 2 + 5.r-4=V(i) 2 -(a;-f) 2 , 



let 



■f = | sin 0. 



174 








The Calculus. 


Then 


dx= 


:f COS 


Bd 




when 


0> V(f) 2 -(*-f) 2 = tcos0; 


when 








a?=4, $=l, 


Therefore 








x=i, 0=-;. 


A = 


it 


f cos 6 


fcos^^-9f 2 cos 2 OdO- 9 " 



II. To Find the Areas into which jr2_j_/2_ a 2 i s Divided by 
* 4 — y 2 (a 2 — * 2 ) =0> and the Area Between the Latter Curve 
and one of its Asymptotes. — The curves are shown in Fig. 53 ; 

x 2 



XI 


p 

afx 

4 




/n 




A5 



Fig. 53. 



they meet where Va 2 — x 2 = 



-vv 



or where x— 



a a 

y=± 



V2' *~~V2 

Of the parts into which the circle 
is divided, the easiest to get is AOD, 
as it is bounded by the graphs of 
only two single-valued functions, 

y x — ya 2 —x 2 above, 

and 

below. 



Va 2 -x 2 



Divide the area AOD in the usual 
way into strips dx wide; the height 
of the strip corresponding to any 
value of x is 



P 1 M->P 2 M=y 1 -y 2 =Vtf 



\fa 2 - 



a 2 -2x 2 
VaF^x 2 



Integration. 175 

area is approximately 

exactly 



and its area is approximately , , ■ dx; the area required is 

Va 2 — x 2 ^ 



J_jl Va 2 - 



X 



Let a;=asin0 ) Va 2 -a: 2 = acos^ 

dx = a cos Odd] a 2 -2x 2 = a 2 (l-2 sin 2 6) 



When x=- -V, 0=--^ 

When s=-^, 0= 4~. 
\/2 4 



AOZ): 



4 a 2 (l-2sm 2 0) _,,, 

— t - a cos (9 <20 



n a cos 

"4 



77 77 77 

= « 2 r cos2^^ = rt 2 j cos<f>idcj>= -^-sm^f =a 2 . 

-a 2 
It is evident from symmetry that the area AC BO is— ^r- — a 2 

= 0.5708a 2 , about. The area between the curve x 4: — y 2 (a 2 — x 2 ) 
= and the asymptote x — a is obtained by summing strips 

2x 2 

high, and is 



Va 2 -x 2 

A r« 2x 2 dx= r 2 2a2 sin2 ^ ^_ jtf^ 

Jo Va^^ 2 Jo 2 

158. Examples. 

1. Trace the ellipse y = ^x—3±i\ / 15-\-2x—x 2 , and find its 
area. Ans. A = 4tt. 

2. Trace the curve y = $x-4:±§ V -x 2 + 20a;-91 and find 
the area enclosed by the curve. Ans. A = 6-rr. 



176 The Calculus. 



a 2 

3. Find the area between ay 2 = x z and y—x. Ans. -y~. 

4. Find the area between y 2 = 3 ax and x 2 = 3ay. Ans. 3a 2 . 

5. Find the area between x 2 (a 2 — y 2 )=a 4: and x 2 = 4:y 2 . 

Ans. a 2 (7r-2). 

6. Find the area between x 2 -\-y 2 = a 2 and 2y 2 = 3ax. 

Ans. ^ 2 (4tt+V3). 



CHAPTEE VI. 

Space Coordinates. 

159. Space Coordinates. — Although no analytic treatment of 
the geometry of three dimensions is to be attempted in this book, 
some of the notation and ideas of the subject will be useful in 
our later work. 

160. Rectangular Coordinates. — A point, P, may be located 
in space by three coordinates, measured as follows : Three 



M' 



M 



Y 



N 



0M~ 



Fig. 54. 



straight lines, OX, OY, and OZ, are given, each perpendicular 
to the other two at their common intersection, 0. These are 
called the coordinate axes, and the three planes determined by 
them, each of which is perpendicular to the others, are called 
the coordinate planes. The distances WP — x, WP — y, L'P — z 
of P from the three coordinate planes, are the coordinates which 
determine the position of P. (See Fig. 54.) 
13 



178 



The Calculus. 



The coordinates are measured from the planes to the point, 
and are positive when directed to the right, forward, or upward, 
and negative in the opposite directions. 

161. Direction Cosines. — The position of any point, P, in 
space may be fixed by giving its distance p from the origin and 
the angles a, p, y made with the axes of x, y, z by the line OP. 
(See Fig. 54.) It is evident from the figure that the relations 
between the rectangular coordinates (x„ y, z) of P and the co- 
ordinates (p, a, /?, y) are 

x—OP cos POL—p cos a, 
y= OP cos POM-p cos /3, 



Since 



OP cos PON= P cosy. 



p 2 = x 2 + y 2 + z 2 = p 2 (cos 2 a + cos 2 p + cos 2 y), 



it is evident that the relation 

COS 2 a + COS 2 p + COS 2 y = 1 

always holds among the coordinates a, /?, y of any point, so that 
(p, a, ft, y) amount to but three coordinates. 

162. Cylindrical Coordinates. — The location of the point P 

may be described by saying that it is 
at the distance z from the x-y plane, 
and directly over the point whose 
rectangular coordinates in that plane 
are (x, y). The point P may also 
be located by giving its distance z 
from the plane XOY and the polar 
coordinates (r, 6) of L', the foot of 
the perpendicular PL' from P to 
Fig. 55. XOY. (See Fig. 55.) 




Space Coordinates. 



179 




163. Spherical Coordinates. — Any point, P, may be located, 
as in Fig. 56, by giving its dis- 
tance p from the origin 0, the 
angle <£ made by OP—p with the 
axis OZ, and the diedral angle 
6 made by the plane ZOP with 
the plane ZOX. This is equiva- 
lent to giving the radius p of the 
sphere centered at the origin 
and passing through P, and then 
locating P on the sphere by giv- 
ing two surface-coordinates <f> 
and 6, analogous to the colati- 
tude and longitude by which a 
location is fixed on the surface 
of the earth. 

The relations between rectangular and spherical coordinates 
are evidently 

x — r cos B — p sin <f> cos 6, 
y—r sin B — p sin <£ sin 6, 
z — p cos <f>. 

164. Equations in Three Dimensions. — If no restriction is put 
upon the values of its coordinates, the point (x, y, z) may, of 
course, occupy any position in space whatever ; if it is given that 
x — a, the point is clearly constrained to move in a plane parallel 
to YOZ, at the distance a from it ; if in addition it is given that 
y — b, the point is further restricted to a line of this plane — the 
line parallel to OZ, at a distance Va 2 + b 2 from it; if, finally, it 
is given that z — c, the position of the point is definitely fixed. 

In any system of space-coordinates, a single relation among 
the coordinates restricts the point to some surface, two relations 
to a curve, the intersection of two surfaces; and three relations 
fix it at the common intersection of three surfaces. The locus of 



180 



The Calculus. 



a single equation in space coordinates is thus a surface, and the 
locus of a pair of equations is a curve. 

The following instances of surfaces are evident from the 
definitions and fundamental theorems of elementary geometry. 

x—y is the plane bisecting the diedral X—OZ — Y. 

x 2 + y 2 + z 2 = a 2 or p — a is the sphere with its center at and 
the radius a. 

y—mx is a plane through OZ, making the angle tan -1 m with 
the plane XOZ , etc. 

165. Cylindrical Surfaces. — An equation in rectangular co- 
ordinates, which lacks one of the coordinates, represents a right 




Fig. 57. 



cylindrical surface of which the elements are parallel to the axis 
corresponding to the missing coordinate. For suppose the equa- 
tion is f{x; y)=0; it is satisfied by the points of the curve 
f(x, y)=0 in the x — y plane, and if this curve, keeping parallel 
to its first position, moves in the direction of OZ, its x and y 
coordinates will not change, so the equation will still be satisfied. 
f(x, y)=0 is thus the equation of the cylindrical surface so 
generated. As an example, consider the cylinder x 2 + y 2 = a 2 in 
Fiff. 57. 



Space Coordinates. 



181 



In the same way an equation f(r, 0)=0 represents a cylin- 
drical surface having elements parallel to OZ. 

166. Analysis of Equations by Plane Sections. — Suppose we 
wish to learn the form of the surface represented by 

.2 „,2 *2 



a 2 + V + c 3 



By making z = Q, we see that its trace on the (x-y) plane is 



the 



r_,, 



ellipse -2 + 4 = 1; similarly, its traces on the other co- 







Fig. 58. 



ordinate planes are ellipses. If we cut it by a plane z — ~k, the 
equation of the section, referred to axes parallel to OX and 
OF, is 



n2 "T" h2 



c 2 -Tc 2 



or 



a 2 ^ b 2 

^r(c 2 -Jc 2 ) -^(c 2 -h 2 ) 



= 1. 



182 The Calculus. 

The section is thus an ellipse with semi-axes 



— Vc 2 -k 2 and -^Vc 2 -& 2 . 
c c 

As h is given a larger and larger value, beginning with zero, this 
ellipse clearly diminishes, vanishing when k = c, and thereafter 
being imaginary. The section is not affected by changing the 
sign of h. 

Sections parallel to the other coordinate planes give similar 
results. The form of the solid, which is called an ellipsoid, is 
now evident. (See Fig. 58, which illustrates any section parallel 
to ZOY.) 

Any equation can be analyzed by these same principles. 

167. Surfaces of Revolution. — An equation in the form 
f(r, z)=0, where r=Vx 2 + y 2 , represents a surface formed by 
revolving about the axis of z the trace of the surface on the x-z 
plane or the trace on the y-z plane. This is because the inter- 
section of f(r, z) =0 and any plane z = h perpendicular to the 
2-axis is given by f(r, ~k) = 0, which when solved will give r=a 
constant, the equation of a circle. Thus x 2 + y 2 = az represents 
the surface formed by revolving about OZ either of the equal 
parabolas, y 2 — az in the y-z plane, or x 2 — az in the x-z plane. 

In the same way, any equation in the form f{r,x)—0, where 
r'='Vy 2 + z 2 represents a surface of revolution about the or-axis, 
and /(/', y)=0, where r"=Vz 2 + x 2 , represents a surface of 
revolution about the y-axis. 

168. Projections of Space Curves. — The orthogonal projection 
of a space curve upon a plane is the intersection with the plane 
of a cylindrical surface containing the curve and having its ele- 
ments perpendicular to the plane. The equation of the projec- 
tion upon one of the coordinate planes is obtained by eliminating 
from the two equations of the curve the corresponding coordinate. 



Space Coordinates. 183 

For instance, the two cylinders 

x 2 + y 2 = a 2 (1) 

and 

y 2 + z 2 = b 2 (2) 

(the first of which has OZ, for its axis, and the second OX) 
determine by their intersection a space curve. The projections 
of this curve upon the planes of x-y and y-z are clearly the 
traces of the two given cylinders. If we subtract so as to elim- 
inate y 2 , the resulting equation 

a*-z*=a*-b* (3) 

represents a third cylinder, with OY for its axis. This cylinder 
contains the curve in question, because the coordinates of any 
point of the curve satisfy the original pair of equations (1) and 
(2) from which (3) is derived and so satisfy (3). Hence the 
trace of (3) on the x-z plane is the projection on this plane of 
the curve. This projection is a pair of straight lines if a — b r 
otherwise a rectangular hyperbola. 

169. Examples. 

Discuss and sketch the surfaces represented by the following 
equations : 

x 2 ip 1 

1. —J + -jo- — 1- Elliptic cylinder, 

2. ^^- + | 2 -=1. Spheroid. 

x 2 -\- y 2 z 2 

3. : — f- -rr =1. Hyperboloid of revolution of one 

sheet. 

4. x2+ f - 4r =0. Asymptotic cone of (3). 



184 The Calculus. 



X 2 y 2 -\- z 2 

5. —2 Tp — = 1. Hyperboloid of revolution of two 

sheets. 

x 2 iP 1 z^ 

6. — 2" + ~fe~ — ~ 2 =1- Hyperboloid of one sheet. 

7. -jo- H — 2~ = — '• Paraboloid (elliptic). 

O C CL 

£2 y2 £± 

8. —5- + T2" + ^r =1- Bull-headed ellipsoid. 

t* c 

Find the projections on the coordinate planes of the following 
curves : 

/ x* + y 2 + z 2 = ±a 2 \ Ans. z 2 + 2ax = ±a 2 on ZOX, 
d ' \x 2 + y 2 = 2ax J z* - ±a 2 (z 2 - y 2 ) =0 on ZOY. 
( x 2 + y 2 = az \ Ans. 2 = 2x on ZOX, 

10 - tz 2 + */ 2 = 2aa; J 2 2 + 4?/ 2 = 4a2 onZOF. 

11. r=a.cos0. Circular cylinder. 

12. 2 = rcota. Cone of revolution. 

13. az — h(a — r). Cone of revolution. 



CHAPTER VII. 
Areas, Volumes, Arcs, and Surfaces. 

170. Areas. — The computation of areas by means of (x, y) 
coordinates has already been explained ; the methods are collected 
here on account of their relation to what follows. 

If the equation of a curve in rectangular coordinates is 
yz=zf(x), y being written as an explicit function of x, the area 
APQB in Fig. 59a is given by 

f ydx, 




an abbreviation for the process of dividing the area into vertical 
strips dx wide and taking the limit of the sum of such strips 
from x — a to x=b as their number is indefinitely increased and 
dx approaches the limit zero. 

If x is given as an explicit function of y, x = cf>(y), the area 
CPQB in Fig. 59b is given by 

•d 

xdy, 

. c 

an abbreviation for a similar process where horizontal strips are 
used as elements of integration. 



186 The Calculus. 

An area bounded by y 1 = f 1 (x) y y 2 =f 2 (x), x = a and x = b is 
given by 

(yi-y 2 )dx, 

Ja 

the limit of the sum of vertical elementary divisions of the area. 
An area bounded by x 1 = cf> 1 (y), x 2 = <fi 2 (y), y — c and y = d is 
similarly 

I (x ± -x 2 )dy. 

When the bounding curve is given by parametric equations, 
these formulas still hold. 

In any definite integral, f(x)-dx, dx is positive if x in- 

Jp 
creases from p to q, negative if x decreases from p to q; where 
f(x) and dx have the same sign, the integration gives a positive 
result ; where they have opposite signs, it gives a negative result. 
In most computations, as for instance in the computation of 
areas, we desire the limit of a sum of the positive values of 
certain elements; for such a purpose the integration should if 
necessary be done in pieces in each of which neither f(x) nor 
dx changes sign, and the order of integration should be so 
chosen in each piece that f(x) and dx have the same sign. (See 
Art 127). 

Whatever the variable x in f(x)- dx may stand for, it may 

Jp 
be used for the abscissa of a graph representing y = f(x), and 

then the value of f(x)-dx will be represented by some area 

Jp 
or sum of areas like that in Fig. 59a. (See Art, 150.) 

171. Examples. 

1. Find the area bounded by y = smx y y=cosx, x — ~ and 

x=^. Ans. 2V2. 

4 



Areas, Volumes, Arcs and Surfaces. 



187 



(> + i) 



2. The strophoid x(x 2 + y 2 ) +a(x 2 — y 2 ) =0 has a loop and a 
vertical asymptote. Find the area of the loop and the area be- 
tween the curve and the asymptote, getting the parts below and 
above the axis of x separately. 

Ans. A 1 = 2a 2 (l— -^) ; A 2 = 2a 2 

3. Find the areas bounded by y = sin x and ^=cos 2x. 

Ans. |V3 and |V3. 

4. Find the area between y 2 = 4 : a(x-\-a) and 27ay 2 = 4:(x — a) 

Ans. |fa 2 (16V2T) 

5. Find the area common to a 8 y 2 = x 4: (a 2 — x 2 ) 3 and 

a s y 2 = x 8 (a 2 -x 2 ). Ans. ~l^-- 

6. Find the area cut off from a loop of x=a<f> — b sin <f>, 
y — a — b cos <f> by the x-axis. (6>a.) 

Ans. (2a 2 + & 2 )cos- 1 4 -3aVF=^. 



172. Sectorial Areas by the y=mx Method. — Parametric 

equations in terms of m— — (see Art. 110) are used chiefly for 

computing sectorial areas bounded by the curve and two lines of 
given slope through the origin. For this purpose special for- 
mulas are better than those already given. 

Let it be required to find the area of the sector bounded by 
the curve x = f(m), y = mf(m) and the 
lines y = m 1 x and y=m 2 x (Fig. 60.) 

Divide the difference (m 2 — m 1 ) into 
any number of parts, each equal to dm, 
and divide the sector into elementary 
sectors by the lines 

y— (m x + dm)x, y— (m 1 -\-2dm)x, 

y=(m 1 + Sdm)x J . . . ., 

y—{m 2 — 2dm)x, y=(m 2 — dm)x. Fig. 60. 

Consider any one of these radial lines, y — mx y reaching to the 




188 



The Calculus. 



point P(x, y), and the next one, y={m-\-dm)x, reaching to the 
point Q(x+Ax, y + Ay.) (Fig. 61.) 

Draw MPS parallel to the y-axis. 

The altitude of the triangle OPS is 
OM=x; the base is PS=MS-MP 
= (m + dm)x—mx=xdm. 

The area of OPS is therefore \x 2 dm. 

The sum of all the triangles inscribed 
in the sector, of which OPS is a type, is 
an approximation to the area of the sec- 




tor, and the area is 



A=i 



m 2 



mi 



x 2 dm. 



(See Art. 154.) 

In the application of this formula x 2 is written as a function 
of m. For the area of a loop having its double point at the 
origin, the bounding lines of the sector are the tangents at the 
origin between which the loop lies, and the sector is the whole 
loop. 

The area bounded by two curves 

and 

z 2 =/ 2 (™), y2 = rnf 2 {m) 

and the lines y=m 1 x and y=m 2 x is similarly seen to be 
4=-jl " (-V 2 — x 2 )dm. 

J mi 

The element of integration in this case is a trapezoid. 

173. Examples. 

1. Find the area of the circle y 2 = 2ax — x 2 by the method of 
Art. 172. 

2. Find the area of the curve y 2 — x 2 {\ — x 2 ) : (a) by the 
integral of ydx, (b) by the integral of \x 2 dm, (c) by substitut- 



Areas, Volumes, Aecs and Surfaces. 189 

ing x=cosd in (a) and changing limits, (d) by substituting 
# = cos0 in the equation of the curve and thence finding para- 
metric equations of the curve, and then integrating ydx expressed 
in terms of 6. Ans. f. 

Find the following areas by the method of Art. 172 : 

3. The loop of y 2 = x 2 -{- x 3 . Ans. -fa. 

4. The curve x* + x 2 f + y 2 -3x 2 = 0. Ans. ^-2V3C 

o 

2 

5. The loop of x z -\-ay 2 — axy = 0. Ans. — . 

6. Between x 3 = a(x 2 + y 2 ) and x=2a. Ans. f-fa 2 . 

7. The loop of the strophoid x(x 2 + y 2 ) +a(x 2 — y 2 ) = 0. 



a 






Ans. _(4-tt). 

8. Between the strophoid and its asymptote. 

Ans. ^-(4+-). 

9. Sector of ellipse #=acos<£, y = b sin <f> between two conju- 
gate diameters. (See example 2, Art. 91.) Ans. — - =— . 



174. Areas by Polar Coordinates. — Problem: To find the 
area of the sector bounded by a 

curve r=f{6), and the lines — a ^_^? 

and 6 — (5. Divide the difference / X\ 

{{3 — a) into any number of parts, / / \\ 

each — dB, and draw the radii vec- / / /\\ 

tores / / \s J \ 

6 = a + d6, = a + 2d6, . . . ., /0k^^^ • 

= /3-dd ^^T^X : 
o " 

to meet the curve. (Fig. 62.) Ym. 62. 

Let P(r, 6) be one of the points of 

division, Q(r+dr, 6 + dO) the next (Fig. 63). Draw a circular 



190 



The Calculus. 



o 




*4*V)* 



Fig. 63. 



arc with the origin as center cutting across the angle dO 

from P. 

This arc is rdO in length, and 
cuts off a circular sector \r 2 dO in 
area. The sum of all the circular 
sectors of which this one is a type 
is an approximation to the area re- 
quired, and the required area is 

A = i[* r 2 d6. (Art. 154.) 

The area bounded by two curves, 

r x =f x ($) and r 2 =f 2 (0), 
and the lines 0=a and $=p is in the same way 

A=i\* (r 2 -r 2 )d0, 

the element of integration being in this case the difference be- 
tween two circular sectors. 



175. 



Examples. 



1. Find the area of the lemniscate r 2 — a 2 cos 26. Ans. a 2 . 

2. Find the area of the cardioid r=a(l — cos 0), or 



r=2asm 2 %0. 



\*tf. • 



Ans. 

3. Find the area of one loop of r—a sin SO. What is the 

7i(l 2 TZO 

total area? Ans. —r=- , — r- • 

12 4 

4. Find the area between the curve and its asymptote, given 
r=2a(sec 6 — cos 6). Ans. Sira 2 . 

5. Find the total area of r=acos0, r=a cos26, r=a cos nO. 

A na 2 .j, . -,, 7ta 2 .» 

Ans. — — , if n is odd ; — =- , if n is even. 



Areas, Volumes, Arcs axd Surfaces. 



191 



6. Find the area of each of the loops of the curve 

r — a cos 0cos 26. 

Ans. 0.3630 a 2 and 0.0148 a 2 . 

7. Find the areas of example 2, Art. 171, and examples 7 and 
8, Art. IT 3, from the polar equation of the strophoid 

r cos 0= a cos 26. 
176. Volumes of Revolution. — If a curve is given by an equa- 
tion that can be solved in the form y — f(x) or in the form 



dx 



k ro 



dx 



i 



m 



(3) 



dx 



~\dy 



-&< 



(4) 



(5) 



Fig. 64. 



l<fy 



3°V 



(6) 



x=f(y), the volume produced by revolving the curve about either 
axis of coordinates, or about any line parallel to either axis, may 
be got by an integration that sums up the volumes produced by 
the revolution about the axis in question of elementary strips of 
the area of the type, ydx, xdy, {y x — y 2 )dx } or (x 1 — x 2 )dy, the 
limits of integration being the extreme values for the solid of 
the variable of integration. 



192 The Calculus. 

In the figures, h, the height of a vertical element, is an ordi- 
nate or a difference of ordinates; I, the length of a horizontal 
element, is an abscissa or a difference of abscissas. 

The axes of revolution are marked with arrow-heads. 

The lengths a and b may be constant or variable; the lengths 
r are always variable. 

The solids generated in (1) and (4) are thin discs; those in 
(2) and (5) are thin discal rings; those in (3) and (6) are thin 
cylindrical shells. 

The elements of volume are : 



(1) 

irll 2 dx 


(2) 
Tr[(a + h) 2 -a 2 ]dx 


(3) 
2irrkdx 


(4) 
irl 2 dy 


(5) 
w[(b + iy-b 2 ]dy 


(6) 

2irrldy 



The forms (1-6) should not be memorized; the elementary 
volume produced should in any problem be got by actual compu- 
tation ; but it should be noted that the volume given for either of 
the thin shells (3) or (6) is the area of its inner curved surface 
multiplied by its thickness dx or dy. The volume of (3) is 

/ dx \ 
actually 2tt( r-\ — — -J hdx, a value intermediate between 2-trrhdx 

and 27r(r + dx)hdx; so the integral of the simpler form gives 
the correct total volume. (See Art. 154.) The other elements 
are volumes of cylinders or the difference between two such 
volumes. 

177. Examples. 

1. Find the volume formed by revolving about the rr-axis the 
area included between the parabolas y 2 — ax and x 2 = ay. 

Ans. y-g-Tra 3 . 



Areas, Volumes, Arcs axd Surfaces. 193 

2. Find the volume formed by revolving the cissoid y 2 (2a — x) 
— x z about its asymptote, using as the element of volume a thin 
cylindrical shell of radius (2a — x). Ans. 2?r 2 a 3 . 

Find the volume formed by revolving the area of each of the 
following about the axis indicated : 

3. (JLV + (jj-\* about the x-axh. Ans. ^hirab 2 . 

4. Area between axes and [-) + (-/-) = 1 about y-axis. 

. ~a 2 b 

Ans. -=-=- . 
15 

5. Area between ay 2 = x 3 and x — y about ?/-axis. 

Ans. -2 2 i 7ra s . 

v 2 x 

6. Area between^- = — and x — a about x — a. 

o 2 a 

Ans. f§-7ra 2 &. 

7. Cycloid x = a(cf> — sin <j>), y = a (1 — cos <f>) about y — a. 

Ans. Large spindle, -^- (3tt + 8) ; small spindle,^- (3tt — 8). 

8. Area between y 2 — -^a(x-\-a) and 27ay 2 = 4c(x — a) 3 about 
a;-axis. Ans. 807ra 3 . 

9. The distance from a point (x, y) to the line x — y is 
±\V2(x — y). Find the volume of the spindle produced by 
revolving the area common to x 2 = ay and y 2 = ax about their 
common chord. (Divide the solid by planes perpendicular to the 

axis of revolution.) Ans. 



30V2 

178. Successive Integrations. — In some problems of integra- 
tion, the value of the element of integration is not immediately 
evident, and is itself determined by an integration. This may 
be done even in the simpler problems that we have already dis- 
cussed, though to no advantage except for a convenient method 
of writing general formulas which is of value in subsequent work. 
14 



194 



The Calculus. 



To explain the process more simply, we will first show how it 
can be applied to some of the familiar problems. 

Let it be required to find the 



O 



% 



*h(*L 



area between two curves, 
yi=/i(aO 



^ 



and 



a 



X 



y 2 =f 2 (x). 



Fig. 65. 



Divide the area into rectangles 
by lines dx apart, parallel to OY, 
and lines dy apart, parallel to OX. (Fig. 65.) Any one of the 
vertical strips is thus divided into small rectangles, each dxdy 
in area, so that the area of the whole strip is dx\dy. In this 
integration dx is the same throughout the strip (is constant), 
and the limits are the least and greatest values of y for the strip ; 
i. e., y x or f ± (x) and y 2 or f 2 (x). As these limits are functions 
of x, the area of the strip, 

f/sCr) 



dx 



A(z) 



dy, 



is itself dx times a function of x. Now if a is the least and b the 
greatest value of x for the area,, the complete area is 

f/»(«) 



r 

Ja 



dx 



fifr) 



dy 



or, as it is commonly more briefly written : 

A— dx\ dy. 

J a )f x (x) 

There is nothing new in this result; | dy is merely f 2 {x) 

J/l(3f) 

—f x (x) or y 2 —y t , so that this is the familiar formula 
A.- \ (V2-yi)dx. 

Ja 



Areas, Volumes, Arcs and Surfaces. 



195 



In the same way, the area bounded by x 1 = F 1 (y) and 
x 2 = F 2 (y) (Fig. 66) is 

2(2/) 



"d CF 2 (y) 

dy dx. 

c " Mm 




Either of these processes may be regarded as 
piling np rectangles to make a strip, then add- 
ing together such strips (each stretching across 
the area) to fill the area approximately, and, 
finally, by indefinitely increasing the number 
of rectangles, obtaining the area. The process 
is briefly referred to as integrating dxdy over p^ 66. 

the area, and is indicated by \\dxdy over the 
area; where \\dxdy is called the double integral of dxdy. 

Suppose again that the area of Fig. 65 or of Fig. 66 is to be 
revolved about the axis of x. Then each element dxdy will 
generate a ring of rectangular cross-section and of inner perim- 
eter 2-rry. Its volume is thus approximately 2-Kydxdy, and the 
total volume generated is 

\\2irydxdy over the generating area. 

In the same way, the volume generated by an area revolving 
about the axis of y is 

^2-n-xdxdy over the generating area. 

The actual volume of one of the generating rings is in the 
first case 

irdx (y + dy) 2 — ndxy 2 = 2irdx ( y + ~- ] dy, 



but in order to establish rigorously the correctness of the inte- 
gration with respect to y, it is merely necessary to observe that 
this volume is intermediate in value between 2irdx(ydy) and 
27rdx(y + dy)dy. (See Art. 154.) 



196 



The Calculus. 



As another example, let it be required to find the area between 
r 1 = f 1 (0) and r 2 — f 2 {B). Divide the area by radii vectores dO 
apart and concentric circles dr apart. (Fig. 67.) 

Any of the small sectors is thus 
divided into figures of a nearly rec- 
tangular shape, a typical one of which 
is bounded by two straight sides dr in 
length and two circular arcs rdO and 
(r + d?*)d9 in length. Its area is very 
nearly rdOdr; the area of the whole 
sector is 




Fig. 67. 
and the whole area is 



J« J Me) 



dO 



rdr 



[W) - 

rdr; 
if lie) 



if a and /3 are the extreme values of 9. More briefly, the area is 
\\rdddr over the area. 

The actual area of the small rectangular division is 
±(r+dr) 2 



ir 2 c 



dr 



\- -)d6dr; 



but as this is intermediate between rdddr and (r+dr)dddr, the 
integral is rigorously correct. (See Art. 154.) 

Completing the first integration of course gives the familiar 
formula 



A=t\[ W-rf)d». 



180. Volumes of Revolution by Polar Coordinates. — In find- 
ing the volume generated by the revolution of a curve given by 



Areas, Volumes, Arcs axd Surfaces. 



197 



its polar equation, successive integrations are actually needed. 
Let it be required to find the volume generated by revolving 
about the initial line the area between r 1 = f 1 (0) and r 2 =f 2 (0). 
Divide the area as in the preceding article and consider any 
elementary division having P(r. 6) for one corner. (See Fig. 
68.) The area of this element is to be taken rdOdr as before. 
As was the case in rectangular coordinates, this area multiplied 
by the distance (2irrsm0) traversed by P gives a sufficiently 
close approximation to the volume of the ring generated by the 
revolution of the elementary division, so that the volume gen- 
erated bv the sector is 






2-irr sin 6 • rd0dr=2ir sin 6d0 



J/l(«) 



and the total volume generated by 
the area is 

7 = 2- T sinOdO 



[* sin 6 dO [ m rHr } 

J* JAW 



if a and (3 are the extreme values 
of 6. 

In the same way, since P , in re- 
volving about the perpendicular to 
the initial line, describes a path 
'2irr cos 6 in length, the volume of 
the elementary ring is 2-nr cos 8 rdQdr, and the total volume is 

2 (0) 




2ir 



f/ 2 ( 
J/i( 



cos 9 dO |" ' r 2 dr 
i(fl) 



More briefly, the volumes formed by revolving an area are : 

2ttJ \r z sin 6 dOdr (revolution about initial line), 
SirJJr 2 cos 6 dOdr (revolution about 1 to initial line), 

the integral in each case being taken over the generating area. 



198 



The Calculus. 



181. Examples. 

1. Find the volume of the solid formed by revolving the 
circle r=2acos0 (a) about the initial line, (b) about the per- 
pendicular to the initial line. Ans. (b) 27r 2 a 3 . 

2. Find the volume generated by the revolution of the cardioid 
r=2asm 2 %0 about the initial line. Ans. f ira 3 . 

3. Find the volume generated by the revolution of the lemnis- 
cate r 2 = a 2 cos 26 about the perpendicular to the initial line. 

Ans. ^ttWVS. 

4. The arc of a cardioid r=2&sin 2 J revolves about the per- 
pendicular to the initial line; find the volume enclosed by the 
outer surface so formed, and the volume of the double spindle 



inside. 



Ans. ^ (16 + 5tt) and ~ (16-5tt). 



5. The area to the right of the perpendicular to the init: 
line between r = a(l — cos 0) and r=a(l + cos6) revolves abc 
this perpendicular, and again, about the initial line. Show th 
the volumes so generated are-f -rr 2 a z and-J-rra 3 . 

6. The lemniscate r 2 = a 2 cos 20 revolves about the initial lin 
Show that it generates the volume 

T Vra 3 V2[3 1og(l + V2)-V2]. 

182. Volumes by Parallel Sections. — The methods that v 
have used for finding the volume of a solid of revolution in re( 
tangular coordinates amount to dividing the solid by planes pei 
pendicular to the axis of revolution, computing the volume of 
cylinder inscribed between two of the planes, and finally inte 
grating to find the limit of the sum of all such cylinders; i. e. 
the total volume. The process is made easy in these cases by thv 
simplicity with which the volume of the typical element can be 
computed (its base being a circle), and is equally easy in any 
case where the area of a section parallel to one of the coordinate 
planes can be simply expressed in terms of its distance from the 
plane. 

If the area. A, of a section of a given solid by a plane parallel 
to the coordinate plane YOZ at a distance x from YOZ is a 



Areas, Volumes, Arcs and Surfaces. 



199 



?iven function of x[A=f(x)'], the volume of the elementary 
iylinder between two planes at distances x and x+dx is 

Adx or f{x)- dx, 

and the total volume is the limit of the sum of all such cylinders, 



or 



\*f(x)dx J 



■I 




Fig. 69. 



\ and x 2 being the minimum and maximum values of x for the 

solid. 
If the area cut by a plane parallel to XOY or to ZOX can be 

expressed in terms of z or y respectively, the volume may be 
^found by a similar integration with respect to z or y. 
!• The section of 



+ 



W~" 



1 



made by a plane parallel to YOZ at a distance x from YOZ is 
an ellipse, the semi-axes of which are the value of z when y=0 
and the value of y when 2 = obtained from the equation of the 



200 The Calculus. 

ellipsoid, or 

— Vfl 2 -r and ^VW^tf. 
a a 

The area of the section is thus 

a 2 v 7 

In the same way, a section of this ellipsoid by a plane parallel 
to XOY at a distance z from XOF is 

and of a section parallel to ZOX at a distance ?/ from ZOX is 
Thus the volume of the ellipsoid is 

( a izhc 
~(a 2 -x 2 )dx 
o a 2 

or 



2) ™(V-f)dy 



'0 

or 

JO c 

i. e., 

183. Volumes by Successive Integrations. — The method of 
parallel sections can be extended to cases in which the area of a 
typical section is not immediately evident, for this area can in 
any case be found by a preliminary integration. If the pre- 
liminary process is made a double integration, the complete solu- 



Areas, Volumes, Arcs and Surfaces. 201 

tion of the problem will necessitate three successive integrations, 
and may be expressed briefly by 

j \\dxdydz throughout the volume, 

where the symbol \ \ \ is read triple integral. 

For the ellipsoid in the preceding article, the process might be 



7 = 8 Jo Mo Mo ' 

which would be described as building up, out of the small ele- 
ments, each (dxdydz) in volume, a column to reach from XOY 
to the surface of the ellipsoid, adding such columns to make a 
slice parallel to YOZ, reaching from XOZ, where y=0, to the 

/ x? 
trace of the ellipsoid on XOY, where y = bJ 1— — (z being 

zero), and finally adding all such slices to build up the solid 
from YOZ, where x = 0, to the end, where x — a. 

As x and y have constant values for the whole column, and x 
the same value for the whole slice, these quantities are treated 
as constants in the corresponding integrations. The integral is 
thus reduced as follows : 



Jo 


r b \/i- 

dx 

Jo 


a* 


V>-* 


y 2 

b 2 


dy 










[1 = 


/.- 


x 2 
' a 2 


s'mO 


Ca 

= 8 

Jo 


7T 

(2 

dx • c 

Jo 


'(• 


- ^) cos 2 6 d6, 
a 2 j 






= 2-bc 


C('- 


x 2 \ 
a?) 


dx = ^7rabc. 









Advantage should always be taken of knowledge already 
gained, so as to reduce the number of integrations as much as 



202 The Calculus. 

possible. The more elaborate process is shown here because the 
notation will be convenient later in many connections, and the 
process in a few. 

184. Examples. 

1. Find the volume between the vertex and the plane x=a..oi 
the elliptical paraboloid 

y 2 z 2 x 
~W + ~tf ~~a' 

by taking an element between two planes parallel to YOZ. 

A -rcabc 

Ans. -^-. 

2. Find the volume cut from a right circular cylinder of 
radius a, by a plane passing through a tangent to the base and 
making an angle a with its plane. Ans. ira? tan a. 

3. What is the volume in example 2, if the plane passes 
through the center of the base of the cylinder? 

Ans. fa 3 tan a. 

4. An isosceles triangle of constant altitude c has for its base 
a double ordinate of the circle x 2 + y 2 — a 2 , and its plane is per- 
pendicular to the plane of the circle. Find the volume of the 

conoid generated as it moves across the circle. Ans. — ^— . 

5. Find the volume of a conoid generated as in example 4, 
except that the triangle moves parallel to the y-axis of the ellipse 

J*. j_ J/L=i 
a 2 + b 2 ' 

and also when it moves parallel to the #-axis. 

A r.abc £ , 

Ans. -^r— tor each. 

6. Find the volume common to two equal right cylinders of 
radius a which intersect at right angles. What is the common 
volume when the two axes are inclined at the angle a ? 

Ans. -^-a 3 esc a. 



Areas, Volumes, Arcs and Surfaces. 



203 



7. A cylindrical hole, diameter 2b, is bored out of a sphere, 
diameter 2a; the axis of the cylinder is a diameter of the sphere. 



Find the volume left. 



Ans. 



¥(«* 



■)i. 



185. Cylindrical Volumes. — The volume enclosed by a cylin- 
drical surface and secant surfaces can be found by the method of 
Art. 183, which can moreover be readily extended so that the 
equation of the cylinder may be used in cylindrical coordinates. 
(See Fig. 70.) 




For instance, the volume cut out from the sphere x 2 + y 2 -\-z 2 
— a 2 by the cylinder y 2 = ax — x 2 , two opposite elements of which 
are a diameter and a tangent of the sphere, may be obtained by 
the integration, 

Ia Wax— x 2 .— -_ 

ax va 2 — x 2 — y 2 dy. 



Here we build up a column z = yW—x 2 — y 2 in height on the 
element dxdy of the base of the cylinder to reach from XOY to 



204 The Calculus. 

the sphere, integrate for y to form a slice across the cylinder, 
and finally sum these slices to form the complete volume. 

We can as well take r — a cos as the equation of the cylinder, 
and r 2 + z 2 = a 2 as the equation of the sphere. Then, building up 
a column of height z— \/ a 2 — r 2 on the element rdOdr of the base 
of the cylinder, integrating for r to determine the wedge from 
the 2-axis across the cylinder, and for 6 to sum up all these 
wedges, we have for the volume : 

IT 

f2" fa cos e , 

F = 4 dO ^Ja 2 -v 2 -rdr=%a\\-\). 

The same result can be obtained from the form in rectangular 
coordinates, but not so readily. 

186. Examples. 

1. Find the volume of the sphere z 2 + r 2 = a 2 , using c}dindrical 
coordinates. 

2. Find the volume cut from the sphere of example 1 by the 
cylinder r 2 = a 2 cos 2</>. Ans. f a 3 (20-16A/2~+3^). 

3. Find the volume common to a right cone the altitude of 
which is h and the radius of whose base is a, and a right cylinder 
having the radius of the cone for its diameter. 

Ans. ha 2 (%-±). 

4. Find the volume of the cylinder included between the 
plane mx + ny + c = z, and the plane of xy, the equation of the 

x 2 v 
cylinder being -y + -p- = 1. Ans. irabc. 

5. Find the volume cut from the cylinder y 2 = 2ax — x 2 by the 
paraboloid x 2 + y 2 = az and the x-y plane. Ans. f 7ra 3 . 

187. In the following figure are collected the most important 
elements of area and volume with all the dimensions infini- 
tesimal. 



Areas, Volumes, Arcs axd Surfaces. 



205 







■ . 






/ 


/ 


i 

dz 


°v 




i | 


< 


A 


» 








m 

! x 





dA 


dx 
*dx.dy 


dA-rd&dr 


"~ Jx 

dV~ dx.c/y.dp 




dV = rd$ - dr ■ dz 



A' pcos Odp 
dV=f>cos G dp. /Ode. dp 



Fig. 71. 



188. Length of Arc. — If s is the length of the arc of a curve 
from any fixed point to any variable point, ds may be expressed 
in some such form as F(x)dx, F(y)dy, F(cf>)d<p, F(6)dd, etc., so 
that we have directly (see Art. 124) : 

The length of the arc of a curve from the point A to the point 

ds, where the limits A and B are the values corre- 
sponding to the points A and B of the variable in the integrand. 



206 



The Calculus. 



Thus the circle of radius a, having its center at the origin may- 
be represented by x 2 + y 2 = a 2 , by x=a cos <f> and y=a sin <f>, or by 
r=a; and we have as corresponding expressions for ds: 

adx ady 



ds= ± 



■Va 2 -x 2 ' 



ad<f> or ±ad$. 



ya 2 -y 2 ? 

The length of the quadrant between the points represented in 
rectangular coordinates by (a, 0) and (0, a) is 

}a Va 2 -x 2 Jo \/a 2 -y 2 Jo r Jo 2 

[Each of the integrands is positive, since from (a, 0) to (0, a), 
x decreases and all the other coordinates increase.] 

189. Surfaces of Revolution. — The surface 8, generated by 
the arc between two points, A and B, of a curve when the curve 
revolves about either coordinate axis, is readily found. This 
surface is the limit of the surface generated by a broken line 
inscribed in the arc as the chords of which it is composed increase 
indefinitely in number and decrease indefinitely in size. 

The area generated by the chord from (x, y) to (x + Ax, 
y + Ay) as it revolves about the a;-axis 
is by elementary geometry 

AS = 2* (y + -^L] V(A^; + (Ay) 2 ; 

hence the differential of the surface is 

dS = 2iry • V(dx) 2 +(dy) 2 = Z-rryds. 

In the same way, if the curve re- 
volves about the axis of y, 




dS=27rxV(dx) 2 +(dy) 2 = %-nxds. 

Hence the surfaces formed by revolv- 
ing the arc from A to B of a given 



Areas, Volumes, Arcs and Surfaces. 



207 



curve about the coordinate axes are 

S=2tt . yds (revolution about OX, or the initial line), 

8=2w\ xds (revolution about OY, or the JL to the initial line). 

If the equation of the curve is more convenient in polar co- 
ordinates, x may be replaced by r cos 9, y by r sin 0. 

190. Cylindrical Surfaces. — To find the area of a cylindrical 
surface included between two secant surfaces, let the axis of z 
be parallel to the elements of the cylindrical surface, so that the 
equation of this surface is in the form 

f(x,y)=0. 

The problem will of course be solved if we find the area of the 
part of the cylindrical surface between either secant surface and 
the x-y plane. Let the equation of this secant surface be 
*=F(x, y). 

Call the section of the cylindrical surface by the x-y plane 
the base; then the equation of the base is f(x t y)=0. (See 
Art. 165.) 

Divide the perimeter of the base into elements of arc each 
equal to ds, and through the points of division draw elements of 
the cylinder, thus dividing the required area into strips. (See 
Fig. 73.) 

Suppose the cylindrical surface to be cut along an element 
and developed on a plane : the base thus 
becomes a straight line which may be 
regarded as an axis of abscissas divided 
into parts ds in length, and the space 
curve in which the cylinder and the 
secant surface intersect becomes a plane 
curve of which the ordinates are the 
^-coordinates of the space curve. The 
required area becomes an area between a Fig. 73. 




208 The Calculus. 

curve, the axis of abscissas, and two ordinates, divided into strips 
in the usual way, so that its value is 

8 = ^zds, 

In this integration, z is the ^-coordinate of the curve 



U=F(x,y)\ 
\f{x,y)=o] 



and so may be expressed as a function of x by eliminating y 
from the two equations of the curve; s is the length of the arc 
of the base, so that ds may be expressed in terms of x through the 
-equation of the base, f(x, y) —0. The integral may therefore be 
put in the form \<$>{x)- dx. In the same way, it may be put in 
the form \0(y)- dy. The limits are the same that would be used 
in finding the perimeter of the base. 

Again, if the surfaces have convenient equations in cylindrical 
coordinates, f(r, 0)-—0 for the cylinder, z = F(r, 0) for the 
secant surface, the same form 

S=$zds. 

can be similarly reduced to the integral of d6 times a function 
of0. 

191. Examples. 

1. Find the length of the arc of y 2 = Sx from (2, -4) to 

(8,8). Ans. 4V~5 + 2V2+log|±0|. 

2. Find the length of a quadrant of the circle x 2 + y 2 = a 2 by 

ca I fdy\ 2 
the form s — \ a/ ' 1 + 1 -— J • dx, by direct integration. 

3. Find the length of one branch of the cycloid 

x = a(cf> — sin <j>), y = a(l — cos<f>). 

Ans. 8a. 

4. Find the length of the catenary V—~k (e x/c -\-e- x/c ) from 
x—— a to x = a. Ans. c(e a/c — e~ a,c ). 



Areas, Volumes, Arcs and Surfaces. 209 

5. Find the length of ay 2 = x s between x=0 and x=5a. 

Ans. "Vr- a. 

6. Find the total area of the sphere formed by revolving 
x 2 +y 2 = a 2 about the #-axis. 

7. Find the surface of the spindle formed by revolving the 
area between a semicircle, the tangents at the extremities of its 
diameter, and a perpendicular tangent, about the latter as an 
axis. Ans. %ira 2 (ir — 2). 

8. Find the surface formed by revolving a branch of the 
cycloid about its base. Ans. --^ -n-a 2 . 

9. Find the surface formed by revolving one branch of the 
curve x=a(<f> — sin </>), y = a(l + co$ <f>) about the #-axis. 

Ans. 3 -f- -n-a 2 . 

10. The cycloid of example 9 revolves about the line y—a, 
forming a succession of spindles alternately smooth and ridged. 
Find the surface of a spindle of each type. 

Ans. ^wa 2 V^and^ira 2 (V2"-l). 

11. Find the whole length of the cardioid r=a(l — cos 0), and 
the length from the cusp to the highest point. 

Ans. 8a and 2a, 

12. Find the surface of the solid formed by revolving r= 
2a sin 2 { about the initial line. Ans. -^ wa 2 . 

13. Find the surface of the solid formed by revolving r — 
2a sin 2 % about the perpendicular to the initial line: (a) the 
inner surface, (b) the outer surface. 

Ans. ^- 2 (3V2-4) and - 4 / ira 2 \f2. 

14. Find the total surface formed by the revolution of the 
circle r=2a cos about the perpendicular to the initial line. 

Ans. 4ttV. 

15. Find the lateral surfaces of the sections cut from the cylin- 
der in examples 2 and 3 of Art. 184. 

Ans. 2?ra 2 tan a and 2a 2 tan a. 

16. Find the lateral surface of the portion of the right cylin- 
der, having the radius a of a sphere for the diameter of its base, 
which is included within the sphere. Ans. 4a 2 . 

17. Find the lateral surface of the cylinder r = a sin 2 J in- 
cluded within the sphere z 2 + r 2 = a 2 . Ans. f a 2 ( 2 V2 - 1 ) . 

15 



210 The Calculus. 

18. Find the lateral surface of r = 2acos0 included between 
z = and r 2 = az. Ans. 4tt<2 2 . 

192. Other Curved Surfaces. — Let it be required to find the 
area cut out from a surface F t (x,, y, z) — by a surface F 2 (x,y,z) 
= 0. This area lies on ^ = and is bounded by a space curve 
of which the equations are 2^ = and F 2 = 0. If we eliminate 
z from the two equations, getting f(x, y)=0, the new equation 
will represent a surface (cylindrical) also cutting F x = in the 
space curve just mentioned, so we can simplify the problem by 
replacing F 2 = by /=0 in the original statement. Moreover, 
f(x, y) =0 is the equation of the projection on the x-y plane 
of the space curve bounding the required area. (See Art. 168.) 

Divide the plane area bounded by this projection into elements 
infinitesimal in both dimensions (e. g., dxdy or rdOdr), and on 
each element erect a prism by drawing ordinates parallel to the 
z-axis. Let P be any point of XOY within the projection 
f(x, y) = and Q the corresponding point directly above it on 
F x (x, y, z)=0. Let the prism corresponding to P and Q cut 
out the area dS from the plane tangent to 1^ = at Q. The 
desired area will be the limit of the sum of elements of the type 
of dS. Kepresenting the element of the projection by dA, we 
have by elementary geometry : 

d#=secy • dA, 

where y is the inclination to the x-y plane of the plane tangent 
to F x (x, y,z)=0 at Q. 
The required area is then 

S = jsecy dA, 

taken over the area bounded by f(x, y)=0, the projection of 
the area S on the x-y plane. 

The projection of S on either of the other coordinate planes 
can of course be used in the same way. If cylindrical coordinates 
are used, F(r, 0, z)=0 and f(r, 0)=O replace F 1 (x, y, z)=0 
and f{x, y)=0 in the preceding discussion. 



Areas, Volumes, Arcs axd Surfaces. 211 

As an example, consider the area cut out from the sphere 
x 2 + y 2 + z 2 ' = or by the parabolic cylinder z 2 = —a(x — a). In this 
case the secant surface is already a cylinder and can be used 
directly; but if we eliminate z, we get as a new secant surface, 
cutting out the same area, x 2 + y 2 = ax, a circular cylinder, which 
is still simpler. In cylindrical coordinates, the sphere and the 
circular cylinder are r 2 + z 2 = a 2 and r=a cos 6. 

It is geometrically evident that for this sphere, 

a a 



for the circle r = a cos 6, dA=rdO dr, and the required area is 

7T 

„ C~2 Ca cos rf i r 

8 = 4a \ de r^-zs =8tf (>-»)■ 

Jo Jo Va 2 -r 2 v ; 



193. Examples. 

1. The cone z = mr or z — cot a- r is cut by the sphere r 2 + z 2 
— 2ar cos 9. Find the area of the surface cut from one nappe of 
the cone. Ans. -na? sin 3 a. 

2. Show that the area cut from the sphere in example 1 is 
4a 2 (a — sin a cos a). 

3. Show that the surface of the solid bounded by the cylinders 
y 2 -\- z 2 — a 2 and x 2 + z 2 = a 2 is 16a 2 . 

4. Show that the area of each piece cut out from y 2 + z 2 = a 2 

by x 2 + f = l 2 is 4a f J^^dy or 4a P sin" 1 V& 2 -s 2 d 
v Jo V a 2 — y 2 Jo a 

194. The Loxodrome or Rhumb-Line. — Suppose the earth to 
be a sphere a miles in radius, and let L and A be the latitude 
and longitude of a point, M, on its surface. Let A be measured 
westward from the Meridian of Greenwich, and let north latitude 

be positive, south latitude negative, so that </>= - ^ — L is the 

colatitude of M . <f> and A are thus coordinates of M. 



212 



The Calculus. 



Let M(<j>, X) and M' (<f> + A<f>, A+AX) be adjacent points on a 
course which passes through M, making 
the angle a with the meridian. (See Fig. 
74.) Let the parallel of latitude through 
W (radius a sin <f>) meet the meridian 
through M (radius a) at m; then 
Mm = aA<f>, mM' — a sin <j> AX. 




Fig. 74. 



Let I be the distance sailed from any fixed 
point of the course to M, and Al-MW. 

In the limiting form of the triangle MmM', 



Cot a 



= a sec a 



sin <f> dX ' 

For a rhumb-line or loxodrome, a is constant, so that the dis- 
tance from M x ($ x , X ± ) to M 2 ((f> 2 , X 2 ) is 



I— (dl=a sec a c£c/>) =aseca(^,- <f> ± ) 
Integrating the two members of 

= COt adX 



(1) 



d<j> 



sm <£ 
between corresponding limits gives 

P-^^COtaP 2 ^ 



or 



tan^ 2 , 

log — =COta(A 2 



tan 



+. 



K). 



(2) 



The direction of the rhumb-line between any two ports can be 
determined from (2), and the distance from (1) ; or if the dis- 
tance run on a given course from a given position is known, (1) 
will determine the colatitude and (2) the longitude of the posi- 
tion reached. 



Areas, Volumes, Arcs axd Surfaces. 



213 



195. Mercator's Projection of the Sphere. — Suppose the earth 
to be mapped on a terrestrial globe, and a cylinder of revolution 
constructed tangent to this globe along the equator ; and suppose 
each parallel of latitude to be projected upon the cylinder by a 
conical surface having its vertex on the axis of the globe. The 
points of the axis chosen as vertices of the cones which project 
the various parallels will influence the form of the projection, 
but in any case if the cylindrical surface is developed into a 
rectangle, the meridians and parallels will form two sets of 
parallel straight lines, mutually perpendicular. 

Mercator's projection is one of this sort, so designed as to 
show any loxodrome as a straight line, having 
its angle with the meridian unchanged by pro- 
jection. 

Fig. 75 represents a Mercator's projection, 
showing a loxodrome, MjM^ two meridians, 
M 1 E 1 and M 2 E 2 , two parallels of latitude, 
M 1 m 1 and M 2 m 2 , and the equator E ± E 2 . 

If the colatitudes and longitudes of M L and 
M 2 are (6 1} A x ) and (<£ 2 , A 2 ), then 

E ± E 2 =za(\ ± — A 2 ) , 

since lengths along the equator are unchanged. Therefore 

m 2 M 2 =E x E 2 — a(X 1 — X 2 ). 

As the angle a is unaltered, and the loxodrome M ± M. 2 is a 
straight line, 

M x m 2 




But 



cota= ^>k = 

m 2 M 2 a(\ ± — A 2 ) 



cot a 



k 2 -\ ± 



log 



tan 



4>: 



tanA 



214: The Calculus. 

(Art. 194) ; hence 

tan A 

M 1 m 2 = a log — • 

tanA 

Consequently, in a Mercator's projection of the terrestrial 
sphere the distance between the equator and the parallel of which 
the latitude is L is the value of M x m 2 when <f> ± = \ —L, <f> 2 = *, or 
is 

, it 

V = a log ^— j^ =alog cot (^ - |) = a log tan ^ + ^ 

tan 



(i-i) 



If lines are drawn on a Mercator's projection representing 
parallels separated by equal intervals of latitude, the distance 
between them will increase rapidly with the latitude. (The rate 

of increase is -^- = a sec L. ) The formula y — a log tan ' ~ + -^- ] 

is therefore called "the law of increased latitudes for the ter- 
restrial sphere." It is also called " the law of meridional parts 
for the sphere." 

196. The Terrestrial Spheroid. — The earth is much more 
nearly a spheroid (ellipsoid of revolution) than a sphere. 

Fig. 76 represents a meridian with its eccentricity much exag- 
gerated. A = a is the equatorial radius, the normal at M mak- 
ing with a the angle L, the latitude of M. The colatitude is 

x is the radius of the parallel of latitude through M . If e is the 
eccentricity, 



yz=VT-^VaF-x 2 



Hence 



Areas, Volumes, Arcs and Surfaces. 
_ dy__ -xVT 



tan</>=tarLr: 



dx 



VaF^x 



2 > 



sec 2 <j> ■ 



a' — e'x' 



x — 



asincf) 



Vl — e 2 cos 2 <j> 



, _ a(l — e 2 )cos <fr d<}> 
(1 — e 2 cos 2 4>)% 




Fig. 76. 

If the arc is measured in the direction P N A, 

4^ = — sec t— sec <f>, ds = dx sec <£, 
dx 

, _ g(l — e 2 )^<£ 



215 



(1) 



(2) 



197. The loxodrome and the Mereator's projection for the 
terrestrial spheroid can now be treated as for the sphere, x tak- 
ing the place of a sin cf> as the radius of the parallel, and ds tak- 



216 



The Calculus. 



ing the place of adcf> as the element of arc of the meridian. We 
have : 



which, from (1) and (2), is: 



cot a = 



Coto 


(,= 


ds 
xd\ ' 


)> is: 








(i 


-e 2 )d<l> 



and 



or 



(1 — e 2 cos 2 cf>)smcf> dX' 
dl = sec a ds; 
_ a(l — e 2 )seca dcj > 

~ (l-6 2 COS 2 ^>)^ * 



(3) is integrated as follows: 

(l-e 2 )d<f> . ,. 

(1 — e 2 cos 2 <£)sm<£ 



, fl — e 2 sm 2 6 — e 2 cos 2 </> , , 

Xcota= -— o ,x • — r^d<£ 

J (1 — e 2 cos 2 



(1 — e 2 cos 2 </>)sin</> 
dcf> f e 2 sin <£ d<j> 



_[ d<j> f e 2 sin 
— J sin <f> }l — e 2 



cos 2 <f> ' 



<f> , e , _ „ 1 + e cos <£ 



Aoota=Iogtan-^ + T log ^^J - 



Hence 



tan 



+. 



(3) 



(4) 



(x x W„-W 2 + e i nlr (l + ecos^Xl-eeos^) 

(A.-Ajcot.-log— ^- + g log (1 _ <eog ^ )(1 + tfeOB+i) 

2 

Therefore, in the Mereator's projection, 

M 1 m 2 = a(X 2 — X 1 )cota 
becomes, when <f> 1 = ^ — L, <f> 2 = %: 



y = a 



logtang+-|-) + -|- log J 



1 — e sin L 
+ e sin L 



Areas, Volumes, Arcs axd Surfaces. 217 

This is the law of increased latitudes for the spheroid; if we 
introduce an auxiliary angle, 0, such that e sin L — cos 6, we shall 
have 



and 



-, -, 1 — e sin L ■> , * 
* lo ST+^ir7X =logtan - 



y = a[logtan(-£ + —j + e log tan f ] . 



The use of infinite series will enable us to integrate (4), and 
will give a more practical form for 

f e 2 sin <f> dcf> f e 2 cos LdL e , 1 — e sin L 

}l-e 2 cos 2 cf> ~ ~ }l-e 2 sm 2 L ~ 2 g l + esin£' 
(See Art. 206.) 

198. Examples. 

1. Show that if we use denary logarithms, and if A represents 
the difference in longitude in degrees, (</> 2 — <£i)' the difference 
of colatitude in minutes, and D the distance in knots, the rhumb - 
line course and distance on the terrestrial sphere is given by 



COta= ^~ 



; 10 tan^-log 10 tan^i 



where log 10 A = 2.12034. 

2. Find the course and distance by rhumb-line and by great 
circle (assuming the earth a sphere) from San Francisco, 

8 h 9 m 43 s W ? 37 o 47 , 2S „ -^ t() Manil ^ gh 3m 50 s E ^ U o 35/ 35* 

N. Ans. : 

Course. Distance. 

Ehumb-Line W. 12° 37.6' S. 6368.2 

Great Circle W. 28° 14.4' K 6051.0 

3. Find the intervals between the parallels of 0° and 5°, 30° 
and 35°, 60° and 65° on a Mercator^s projection of a sphere 
of radius a, and on a Mercator's projection of a spheroid of 
equatorial radius a, eccentricity e — 0.081697. 



CHAPTEE VIII. 

Series. 

199. In the study of functions it is often the case that the 
development of a function in power-series gives an expression 
more readily handled than the original form. This is particu- 
larly true when we attempt to integrate functions whose integrals 
are very complicated or cannot be expressed at all in terms of 
familiar functions. 

200. Development in Series. — Suppose that a function of x 
can be developed into a power-series 

f (x) = a -\-a 1 x+a 2 x 2 + a 3 x 3 + .... + a n x n -\- . . . . ; 

then, assuming that the method of finding the derivative of a 
finite power-series applies also to an infinite power-series : 

f(x) =a 1 + 2a 2 x+3a s x 2 + 4:a /t x s + +na n x n - x + , 

f(0)=oi; 

f"(x) = 2a 2 + 2 • 3a s x + 3 • 4:d 4 pc? + + (n-l)na n x n - 2 + , 

f'(0)=2a 2 ; 

f"( x )=2-3a 3 + 2'3'4:a>4X+...+ (n-2)(n-l)nanX n - z + . . ., 
f"(0)=2-Za 3 ; 

? 

f( n) (x) =2 • 3-4.. (n — 2) (n — l)na n + multiples of powers of x, 
fW(0)=a n \n. 

Since f(0)=a o , we have Maclaurms Series: 

««)-/(o)+*-f(o)+|-r(o)+|no)+.... 

+ ^P'(0)+.... 

Exercises. 

Use Maclaurin's Series to obtain the series for (a+x) m , 
sin x, cos x, e x and a? already developed in the Trigonometry and 
in the Algebra. 



Series. 219 

201. Practical Methods. — We can develop sec x directly. Since 
csc(0) = oo ? esc a; cannot be thus developed; by noticing that 
esc x — sec (x— 2), we can develop cscz in powers of (x — £). It 
is, however, much simpler to replace cos x and sin x in the for- 
mulas sec x— , esc x— -. by the first few terms of their 

cos x sin x 

developments, and carry out the divisions to determine an equal 
number of terms of sec x and esc x. 

Maelaurnr's Series is used in practice to develop such func- 
tions as it readily applies to ; then the development of any simple 
function of these can be obtained by elementary processes. 

202. Examples. 

1. Write four terms of the development of cos 2x, and thence 
obtain four terms of the development of sin 2 x and of cos 2 x. 



Ans. 



sm 2 ^^- x 4-^-3^ + 

cos 2 x=l — x 2 -\ — =- — ~r=- + . . 



3 45 

2. Find three terms of the development of cscz from three 

1 x 1x^ 
terms of the development of sin x. Ans. — + — + ^7^ . 
r x 6 360 

3. G-iven c. m. 30° = 0.5.236, compute to four decimals (by 
logarithms) the first three terms of esc x, and compare their 
sums with esc 30°. 

4. Find three terms of the development of tan x (from sin x 

and cos x). Ans. #+ — + ^-=- 4- . . . . 

5. Find the developments of i(e x + e- x ) and \{e x — e~ x ) from 
the development for e x . 

/yt2i /y»4 /v*D 

14- — + — 4- — 4- 

1+ |2 ^ [4 ■ [6 + ' ' ' •' 



Ans. 



/Y»3 -|iO jy i 

X~\ TFT ~l~ —^~ ~^~ TW 

li II .11 



220 The Calculus. 



6. Find the development of Vl — cos x, considering Vl — cos x 



x — 

as a function of -=- . Ans. \/2 



JO Jb | Jb 

2 ~2 3 |y w\i[~ 



7. Find the development of cos 3 x from the identity cos 3 x 
= f cosz + |cos Sx. Ans. l—^x 2 + ^x i —^x G + .... 

203. If the derivative of a function is easier to develop than 
the function itself, it should be developed, and the resulting 
series integrated on the assumption that an infinite and a finite 
power-series obey the same laws of calculus. The arbitrary con- 
stant that appears can always be determined from the value of 
the function for some one value of its argument. Often the 
development of the derivative follows from some familiar theo- 
rem. For instance, to develop tan -1 x : 

By the Binomial Theorem, 

-j— (tan -1 ^) = -— j — ~ =1 — x 2 + x* — x G + x 8 — . . . ., 
dx v ; l+x 2 

tan -1 x=$[l — x 2 + x i — x 6 + x 8 — . . . . ]dx 

\ 

/y»3 /yt5 /y*l /y»9 

tan- 1 = (7 = 0, 

/VtO /y»0 /y»7 



204. Examples. 



1. Develop x tan" 1 x — log Vl + x 2 by first developing its de- 
rivativ, Ans. *--£+.*_ ^ + ... , 

2. Given £=4 tan" 1 i-tan" 1 ^ -Han" 1 -^ , find the value 
of 7T to eight figures from the series for tan -1 x. 

Ans. tt=3.1415926. 



Series. 221 

3. Develop sin -1 x by first using the Binomial Theorem to 

develop its derivative. 

A , 1 , . 1.3 x 5 , 1.3.5 x 7 , 

Ans. ^+^3^ 3 +^5+-2^ T +.... 

4. Compute sin -1 J = £ to four figures from the series of ex- 
ample 3. 

205. Approximate Integration. — The method of the preced- 
ing article enables us to find an expression in series for an inte- 
gral if we can develop the integrand, and so is of great advan- 
tage in the evaluation of an integral which is not a known 
function. 

For instance, the length of a quadrant of the ellipse x — 



a cos <f>, y- 

n 

HI 


= aVl — e 2 sin<£ i 


s 

> 


Vl-e 2 cos<£- di 


77 

HI [* " T C ° S2 * - 2?2 C0S4 *~ ¥g e " C0S6 * 


J. * O * D O Q . 

g 4 | 4 «W* .... 


^; 


<?=-f 


" e 2 3e 4 5< 
1 4 43 4 


>e 52 . 7^8 72 . 9 . g io -j 




4 47 4s 


* 


206. Examples. 


1. Show that if a=10, e=0.1, § = 15.669 is the length of an 


elliptic quadrant. 


2. Show $x n e~ x2 dx=x n+1 


~ 1 x 2 a: 4 
_n + l n+3" 1 " (n + 5)| ? 




a; 6 z 8 




(n + 7)_[_3 r (n + 9)|J •"■ 


• 



3. Show that the area in example 4, Art. 193, is 



222 The Calculus. 

if m = - 



a 



**fi -l m * _l ( 1>3 ) 2 (1-3-5) 2 6 . \ 



and if a=2b, 

£ = 1.0357r& 2 . 



e^ , 1 — e sin L 
2 l0g l + esin£ 



4. Show that the second term in the law of increased latitudes 
for the terrestrial spheroid (Art. 197) 

' e 2 cos LdL 
1 — e 2 sin 2 L 

^-e 2 sinL(l + ie 2 sin 2 Z + ie 4 sin 4 L+ ....). 

5. Given that a = 3437.7 minntes of equatorial arc, 

log e N= 2.3.026 log 10 if, e = 0.081697, 
show that in Art. 197 

y= 7915.7 log 10 tan ( 45°+ -4) -23 sin £, approx., 

in minutes of arc. 

6. Show that 'the length of the loxodrome on the terrestrial 
spheroid (Art. 197) is 

Z = a(l-e 2 )seca f ii2 [l+fe 2 sin 2 L+- 1 -g- 5 -e 4 sin 4 L+ .... ~\dL. 

JLi 

207. Differential Equations. — An equation involving deriva- 
tives of a function is called a differential equation. 

Sometimes such an equation is formed for the sake of de- 
veloping a function; as an example, suppose we wish to find the 
law of coefficients for the development of tan x. We let ^=tan x, 
so that 

Jl = sec 2 x=l + y 2 , 
and our differential equation is formed: 



Series. 223 

Since tan( — x) — — tan (a;), no even powers can occur; as- 
sume, therefore, 



= 


a x x 


+ a z x* 


+ 


a b x ° 


+ a 7 x 7 


+ . . . . 










*== 


a*x 


2 


+ 


of* 




+ a. 2 * 10 




+ a. 2 x w 


+ . 








+ 2a 1 a. i x 


t +2a 1 a 5 a 


6 + 2a x a 


jX* + 2a 1 a 9 x 10 + 2a 1 a 11 x 12 + 2o 1 a 13 x 1 * + . 














+ 2a 3«5 


x 8 + 2a 3 a.z l0 + 2a 3 a 9 


z 12 + 2a 3 a 11 x' 


»+ • 


















+ 2a 5 a 7 


z 12 + 2a 5 a 9 x 1 
+ 


* + . 





-^|-=a 1 + 3a3X 2 ^5a 5 z t + 7a 7 a; 6 + 9a 9 x 8 + lla 11 x ,0 + 13a 1 3X 12 + .... 

Comparing coefficients of like powers of a:, we have 

a x = l, 3a z = a x 2 , ha 5 — 2a x a z , 7a 7 = 2a x a 5 + a 2 



3 3 



9a 9 = 2^0^ + 2a 3 a 5 , lla ±1 = 2^9 + 2a 3 a T + a 2 . 
The computation for several terms follows : 





^=1. 


3a 3 = a 1 2 = l. 


a 3 =J. 


5«5 = ^l«3 = f- 


«5=T 2 5. 


fc,-«MiW-i+J- U + 5 -g. 


17 
a7= 5.7-9' 


34 4 344-28 62 


62 


9 ~5-7-9 l ~9-5 _ 5-7-9~5-7-9" 


Oi- 5.7.92 


«-«+*>+£*+ 5 . ¥.,*+..£,, 


.r 9 + . . . . - 



Exercises. 

1. Since sec(O) =1 and sec( — cc) =sec #, assume 
1/ = sec a; = 1 + a 2 x 2 + a^ + # 6 a; 6 + a s £ s + . . 

and show that -^\ = 2i/ 3 — y. 

Substitute, and determine a 2 , a±, a 6 : 

sec x = l+ix 2 + ^x 4 + f^x°+ .... 



224 The Calculus. 

2. Find five terms of the expansion of log cos x. 

An -?- *lL _ ^ _ 17z 8 __ 31x 10 

S ' 2 12 45 2520 14175 

208. Elementary Series. — The following developments of ele- 
mentary functions are collected here for reference: 

(l + x) n = l + nx + n(n-l)~- +n(n-l)(n-2) ^-+ . . . . 
a x = l + xloga+-j^(xhga) 2 +-^- (xloga) 3 + . . . . 



e x z 


= 1 + 


x + 


X 2 

P 


+ ■ 


a; 3 

|3 


+ 


X 4 

14 


+ . 


lqg(l + aO = 


-X — 


x 2 
2 


+ 


z 3 
3 




X 4 

4 


+ 




sina;- 


-X — 


x 3 
13 


+ 


15 


— 


M 


+ 


. . . 



. X 2 X 4 X 6 . 

. , z 3 , 2X 5 17x 7 

1 , x 7x 3 31x 5 , 
CSCa;= ^+T + m + l5l20 + 

61x 6 





sec x- 


= 1 + 


x 2 
2 


- + 


5x 4 i 
24 ' 






cot rr = 


1 

X 


— 


3 " 


x 3 

45 i 


log 


sin#= 


:l0g; 


x — 


X 2 

6 


x 4 
180 


log 


secx = 


X 2 

" 2 


+ 


x 4 
12 


T 6 

+■16 + 



720 

2x_ 5 
945" 



2835 



= —log CSC X. 
= — log COS X. 



logtan^logx+f + ^ + ||L + . . . . = - log cat*. 



Series. 225 



. t , x* 3x 5 5x 7 

113 5 

esc- 1 3= _ + _ + ^ + Y12P + • • • • =I-sec- 1 x. 

/y»3 zyi5 zv*7 

tsjr x x — x 5- + -=- — -=- + . . . . =|— cot -1 a;. 

cot -1 x— -^-* + ^n> — ^-7 + - • • • =\ — tan -1 a;. 

x Sx 3 5x° 7x 7 2 

209. Development when fM (0) = oo.— If f(x), f(x) or any 
higher derivative is infinite when x = 0, f(x) cannot be developed 
by Maclaurin's Series. For instance, an attempt to apply the 
method directly to log x will fail; log(l + a;), however, is readily 
developed. If 



f(x)=log(l + x), 

f(*) = (i+*)-S 

f'(x) = -{l + x)-\ 
f"(x)=2(l+x)-\ 


/(0)=0; 

f(0)=i; 
f'(0) = -l; 

/'"(0)=2; 


/(«)(0) = (-l) n - 1 
log(l + x)=x--?Y ■ 


|»-l(l+a;)-», 

|»-1: 

a 3 x 4 
^ 3 4 

_i)»-i_^L + . . . 

n 



From this development the series actually used for computing 
logarithms can be derived, as in the Algebra, Art. 162. 



log(7i+7i) —\ogn-\-2 



*( 



* \ 3 . 1 ( 



2n + h^ 6 \2n+hJ ^*\2ii + h 



+toTTL +' ' ' 



210. Taylor's Series. — The procedure that was necessary in 
the case of log(l + a;) is useful in many connections. Suppose 
we wish to compare a particular value of f(x), say f(x ), with 
16 



226 The Calculus. 

adjacent values. All values of f(x) may be represented by 
f(x +z), where z is variable, and these values will be nearer to 
the value of f(x ) as z diminishes. To develop f(x + z), which 
is a function of z (x being a constant), we assume 

f(x + z) = a + a 1 z + a 2 z 2 + a t z 3 + . . . . +a n z n + .... 

Now, since x is a constant, d(x + z) =dz; hence 

and similarly for the higher derivatives. Consequently, if we 
take successive ^-derivatives of both members of the assumed 
identity, and subsequently make z = in each of the resulting 
identities, we have : 

f(x +z) =a 1 -\-2a 2 z-\-3a 3 z 2 + . . 

f"(x + z) =2a 2 + 3 • 2a s z + + n(n-l)a n z n - 2 + , 

f"(x + z) = \S • a 3 + . . . . + n(n-l) (n-2)a n z n ~ 3 + . . . . , 
? 

f{x )=%, f(x )=a 19 f'(x )=2a 2 , 

f"(so)-={3-a* ...., / (n) (^o) = K •••> 

so that 

This development is called Taylors Series; it is often spoken 
of as the development of the function f(x) in the neighborhood 
of the value x of its argument x. Maclaurin's Series is a special 
case of Taylor's, the development in the neighborhood of the 
value zero. 



Series. 227 

Taylor's Series is especially adapted to the study of the in- 
crease of a function of x when x is increased from a particular 
value x by the difference or increment dx. 

Call f(x)=y, f(x )=y , f(x + dx)=y + Ay; then 
y + Ay=f(x + dx) =f(x 6 ) + dxf(x ) + <*£ /"(*,) 

211. Finite Differences. — The development 
Ay=dx ■ f(x ) + J*2 f'K) + ^- f'K) +• • • • 

+ W! / (.) (a . o)+ .... 

gives, to any required degree of approximation, the increase in 
the function y caused by the increase dx in its argument, The 
number of terms needed will depend on the nature of the func- 
tion, the size of dx, and the accuracy required. In many practical 
cases, one term is sufficient; the resulting formula, 

Ay — dx- f(x ) or Ay = dy, 

is merely the assumption that the increment of y and the dif- 
ferential of y do not differ in the decimal places that it is desired 
to have correct. Graphically this means that the curve y = f(x) 
and the tangent to it at (x , y ) have ordinates differing by a 
negligible amount when x — x Q -\-dx. The second term, 

will generally show whether this difference is really negligible. 

In any case, enough terms of the Taylor's Series are computed 
so that the last one does not affect the decimal places retained in 
the computation, and generally one or two more places are re- 
tained in the computation than are desired in the result. 



228 The Calculus. 

For instance, if y=log 10 x, 

Xq & Xq O Xq 

v ' n V 
or 

. [dx 1 (dx\ 2 , 1 tdx\ 3 

X-iy-i /dxY 

n \xj _ 

(/x = log 10 e = 0.43429.) 

If dx is positive, this is an alternating series, so that the error 

committed by taking Ay—dy— — is less than -~ I J , 

and the value of Ay so taken is too large. 

212. Examples. 

1. If it is assumed that log 10 (N+n) =log 10 N+ -=- , how 

large a fractional part of N may n be if the error in the assump- 
tion is not to affect the fifth place in the mantissa? [Error 
< 0.000005.] How many figures of ^ will be useful in the 
approximate computation ? 

Ans. (4rr) <-00480; three figures; use fi = 0.434. 
log 10 (N + n)=log 10 N+'-^. 

2. Find by Taylor's Series the value to 5 decimals of sin 
(30° 30') from the functions of 30°. 

3. Find the increase in log 10 sinz, when £ = 30°, dx=l\ to 7 
decimals, and check by tabular difference between log 10 sin 30° 
and log 10 sin 30° 1'. 

4. Find the increase in log 10 tana:, x = 45°, dx — V. 



Series. 229 

5. Develop y=log 10 sec x in the neighborhood of the value 
x = tan -1 f and thence find to seven decimals the values of Ay 
when dx—V, dx—\# and dx=l° 40'. 

Ans. 0.0000948, 0.0009504, 0.0097663. 

6. Develop y = sin 2 x in the neighborhood of £ = 45°, and 
thence find to five decimals the values of sin 2 46°, sin 2 50°, 
sin 2 60°, and sin 2 75°. 

Ans. 0.51745, 0.58683, 0.75000, 0.93301. 

213. Small Changes in the Astronomical Triangle. — The for- 
mula Ay — dy gives valuable approximations to the changes pro- 
duced by small variations in the parts of the astronomical tri- 
angle. Of the five parts, L, d, h, t, and Z , three must be given 
to determine the triangle, and then the other two can be found. 
Each of the other two is thus a function of the three given. For 
instance, any one of the four parts, L, d, h, t, can be expressed 
as a function of the other three by means of the equation 

sin/z, = sin d sinL + cos d cos Lcos t. (1) 

The results obtained from (1) can be simplified by means of the 

equations 

cos h sin Z = cos d sin t, (2) 

cos h cos Z — cos L sin d — sin L cos d cos t. (3) 

For instance, suppose L, d, h are given, and it is desired to 
find the effect on t of small errors in the data. We will suppose 
each of the given parts in turn to vary, the other two being con- 
stant, and thus find three errors produced in t, the sum of which 
will be the total approximate variation of t. 

First, let L and d be constant, h and t variable; then, dif- 
ferentiating (1) we find: 

cos 7w?7i = + cos d cos L(— sin tdt), (V) 

or, by (2) : 

cos lidh= —cos L sin Z cos hdt, 
dt= —esc Z sec Ldh. 



230 The Calculus. 

Next, let d and h be constant, L and t variable, 

= sin d cos LdL — cos d cos t sin LdL — cos ^ cos L sin W#. 

Then by (3) : 

cos d cos L sin tfj£= (sin cZ cos L — cos ^ sin L cos tf)^L 
= cos h cos Z<£L, 
or, by (2) : 

j t • j-jj ry sin t COS <2 7r 

cos a cos iv sm tat — cos Z : — = — aL, 

sinZ 

dt = cot Z sec LdL. 

In like manner, differentiating (1) on the assumption that 
t and d are the only variables, and reducing the result by means 
of 

cos h cos ikf =cos d sin L — sin d cos L cos £ (4) 

and 

cos L sin tf = cos h sin If, ( 5 ) 

we find 

dt= cot M seed' dd. 

Finally, if L, d, h are given, subject to errors AL, Ad, All, the 
error in t is approximately 

At = cot Z sec LAL + cot M sec dAd — esc Z sec LAfc. 

In the same way, if d, t, h are given, subject to errors Ad, At, 
Ah, the error in L is approximately 

AL= —cos M sec ZAd+tan. Z cos LA£+sec ZAh. 

In the navigation problems of the time-sight and the latitude- 
sight, the error in declination is negligible, and these formulas 
become, for the time sight, 

At = cot Z sec LAL — esc Z sec LAh 
and for the latitude-sight, 

A£:=tan Z cos LAt + sec ZAh. 



Series. 231 

In either of these, the error Ah arises from inaccuracy in ob- 
servation. In the time-sight, AL is the error in the assumed 
(dead-reckoning) latitude. If the time-sight is taken when the 
sun is near the prime-vertical, Z is nearly 90°, cotZ and esc Z 
are small, and the error in longitude, At = sec L(cotZAL 
— esc ZAh), is small, so that the longitude found from the obser- 
vation is more accurate than the dead-reckoning longitude. The 
latitude-sight, on the other hand, should be taken when the sun 
is near the meridian, for then Z is nearly 0, and tan Z and sec Z 
are small. The effect of the error in the dead-reckoning longi- 
tude and the error in h are thus both made small, so that the 
latitude found is more accurate than the dead-reckoning latitude. 

Formulas involving Z in place of t can be derived in the same 
way from 

sin d—smh sin L + cos ft cos L cos Z. (6) 

214. Examples. 

1. The time and amplitude of sunrise or sunset, computed on 
the assumption that ft = 0°, are given by 

sin A = sin d sec L, cos t — — tan d tan L. 
Supposing the assumed values of d and L to be accurate, show 
that the change Aft = — 50', due to the mean semi-diameter of the 
sun (16') and the mean refraction (34'), will cause a change, 
At = sec A secL • 50'. 

2. Show that, at the Naval Academy (Lat. 38° 58' 53" K) 
the apparent solar times of sunrise and sunset are about as 
follows : 

Winter Solstice, Equinoxes, Summer Solstice, 

Dec. 21. Mar. 21, Sept. 21. June 21. 

d=23° 27' 07" S. d=0. d=23° 27' 07'' N. 

, A . . A. A 



hrs. min. hrs. min. hrs. min. 

Sunrises... 7 17.2 a.m. 5 55.7 a.m. 4 32.8 a.m. 



Sun sets 4 42.8 p.m. 6 04.3 p.m. 7 27.2 p.m. 

aat the error made in computing h f 
, L is 

Ah = cos ZAL — sin Z cos LAt approx., 



3. Show that the error made in computing h from assumed 
values of t, d, L is 



232 The Calculus. 

if Ad = 0. It follows from this that a line of position can always 
be laid off by St. Hilaire's method on a large scale chart, regard- 
less of the azimuth of the sun. 
4. Show that, if Ad = 0, 

AZ = tan h sin ZAL — cos M sin Z esc tM. 

Use sin t cot Z — cos L tan d — sin L cos tf, and simplify the results 
by using cos t cos Z — sin L sin £ sin Z-— cos If, and the for- 
mulas (1) and (2) already given. 

215. Simpson's Rules. — When an area or a volume is to be 
found and the equations of the bounding curves or surfaces are 
not known, it is possible to determine the desired result or a fair 
approximation to it from measurements of a number of ordinates. 
More generally, when a definite integral is to be found, it is pos- 
sible to express it, sometimes exactly, more often approximately, 
in terms of a few values of the integrand, even if the form of the 
integrand is not known. 

The problem in this method is: To find 1 f(x)- dx } given the 

]a 

values f(x ), f(x t ), f(x 2 ), . . . ., f{x n ) of f(x) for a number 
of values of x uniformly distributed in the range from a to ~b. 

216. Simpson's First Rule. — Suppose three values of the un- 
known function to be given; shift the origin so that the middle 
one corresponds to x=0, and call b — a=H. In this case we wish 

H 

to find f 2 f(x)-dx, given /(-f), /(0), /(f). For the sake 

J "f 

of brevity, call these given values y ± , y 2 , y z respectively. 

Assume the function to be of the form 

f(x)=A+Bx + Cx 2 + Dx 3 ; 
then 

./ H\ A BH CE 2 DH 3 n . 

y ^ f {- -2) =A ~ -T + ir-^"' (1) 

y 2 =f(0)=A; (2) 



Series. 



233 



./ H \ A x BE- 

y*=f[-2~) = A + -o~ + 



CIP DIP 

4 + 8 



(3) 



If, in Fig. 77, the heavy line 'PQRS is the graph of f(x), the 
unknown function, and the dotted line pQRS is the graph of 
y = A+Bx+Cx 2 + Dx 3 , the true integral and the approximate 
integral we are about to find are represented by the areas 
AQRSBA, the first bounded by the full line, the second by the 
dotted line. The function A + Bx + Cx 2 + Dx 3 and its graph are 
of sufficient flexibility, even with the three values at Q, R, S 
fixed, to conform pretty closely to any given function and its 
graph. 



<? 


/>' 




VgS?/ 


,4 


jf 


. # . 




/ >. 




A 


9 



'. , Bx 2 , Cx 3 , Dx ( 



Fig. 77. 
We have for the approximate integral : 

H 

J H (A+Bx + Cx 2 +Dx*)dx 

Adding relations (1) and (3) above, 
«, . CH 2 



so 





— 1 g -•/! 


T# 3 > 


^ — </2> 


CH 2 
2 


=yi+ys-^ 


C# 2 

12 


6 



234 The Calculus. 

hence 

H 

J 2 H (A+Bx+C*°+D*)dx=H U+ C ^)=«(y 1+ ±y 2 +y 3 ). 
Therefore, approximately, 

H 

\*_ B m- dx= f [y, + ±y 2 +y s ] = f [/(-f)+4/(0) +/(f)]. 

2 

The position of the origin relatively to a and b has no effect 
on the value of the integral; it has merely been necessary for a 
definite discussion to fix it somewhere. As H represents (b — a), 

and the value of x midway between a and b is ~^ , 

" a f {x ).d x =^±[f(a)+4f p±$) +/(&)] 

approximately, or 

cb b — a 

j ^ f(x)- dx = -g- [yx+^+ysJ, 

where y 1? y 2 , 2/ 3 are the values of f(x) at the beginning, the 
middle, and the end of the range from a to b. This formula is 
known as Simpson's First Rule. 

The assumption which makes the result approximate is that 
f(x) is a polynomial of the third or lower degree; if this is really 
so, the integration is exact. 

217. Examples. 

1. If two perpendiculars, y ± and y s in length, a distance h 
apart, are dropped from points of a curve to a straight line, and 
a third perpendicular, y 2 in length, is drawn midway between 
them from a point of the curve to the straight line, show (a) 
that the area bounded by the curve, the straight line, and the 
perpendiculars y x and y 3 is approximately 

4 = -g-(2/i+42/ 2 +# 3 ), 



Series. 235 

and (b) that the volume produced by revolving this area about 
the straight line is approximately 

2. Find the following integrals by Simpson's First Eule : 

[ 2 (2 + x + 2x 2 + 3x 3 )dx=^- 
and 

P (7x 3 + 3x 2 + 2x + 5)dx = 12. 

Find the following volumes by Simpson's First Eule : 

3. The volume of a sphere. 

4. The volume formed by revolving about y = the segment 
from x=0 to x = a of the parabola ay 2 = b 2 x. Ans. ^-n-ab 2 . 

5. The volume of a barrel formed by the revolution about the 

x 2 v 2 
major axis of the segment of the ellipse ^ + -— = 1 between the 

ordinates through its foci. Ans. 56.64?r. 

6. The volume of a barrel formed by the revolution about the 
major axis of the segment between the ordinates through its 
foci, if the length of the barrel is 2h and the diameter at the 

bung is 2b. Ans. %irb 2 h /2 , 72 • 

7. The volume of a spherical segment, altitude Ji, radii of 
bases, b and c. Ans. fah{Zl 2 + Z(? + li 2 ). 

8. The volume of a capstan in the form of a hyperboloid of 
revolution, each base of radius b, circle of gorge of radius g, 

altitude h. Ans. ^(b 2 + 2g 2 ). 

o 

9. The volume of a conoid of height h having a circle of 
radius a as base. Ans. -Jthz 2 /*. 

218. Closer Approximations. — When the approximation of 
A+Bx+Cx 2 +Dx* to f(x) is not sufficiently exact, there are two 



ways of coming closer to the true value of f(x)-dx, 

Ja 



one is 



236 The Calculus. 

to break up the range into equal parts, and to apply the method 
just described to each part; the other is to assume f(x) equal to 
a polynomial of higher degree and to determine correspondingly 
more values of f(x) from which to compute the integral. Either 
method involves an increased number of measurements of f(x), 
or of computations of f(x), for cases in which f(x), though not 
integrated, is known. 

If we follow the first method, dividing the range (b — a) into 
two equal parts, and applying the method to each of them, 

t/i> y*, 2/3; y*> y 5 , 



being the values of f(x) at the beginning, the points of quadri- 
section, and the end, we have 



p*3 



i(h-a) 



J( x )dx=^^- (y 1 + 4,y 2 + y 3 ), 

b J(x)dx= *( h - a> > (y 3 +4y 4 +y 5 ), 

b b — a 

a f(x)dx = -j2-(y 1 + fy 2 + 2y B + fy i + y 5 ). 

In the same way, if the range (b — a) is divided into any even 
number n of parts and y 19 y 2i y 3 , . . . . , y n are the values of the 
function at the beginning, the points of division, and the end, 

J f(x)dx=-^ (y ± + 4:y 2 + 2y 3 + 4:yi + 2y 5 + .... +#„ +1 ). 

In the parenthesis, the first coefficient is 1, the last is 1, the 

72 

others are alternately 4 and 2, beginning with 4. There are -5- 
4's, 2^-2% and 2 l ? s: 



4 (D +2 (^) +3(1)=3)l - 



Series. 237 

The sum of the coefficients is three times the number of divisions 
of the range. The rule is often abbreviated : 

r. "*>*= 1+4+2+"'?.. +1 (i > 4 > 2 > ■ ■ • ■> *>• 

The simpler rule is similarly written : 

O^ifi+T^ 4 ' 1 )- 

219. Examples. 

1. The segment from x = to x = a of ay 2 = b 2 x revolves about 
x~a. Apply Simpson's First Eule, dividing the solid by planes 
perpendicular to the axis of revolution, first, \ apart, second, | 
apart; and show that the errors are fa and l 1 24 of the correct 
volume, y|-7r<z 2 &. 

2. Treat as in example 1 the volume formed by revolving the 
same area about the tangent at the vertex, and show that the 



f 3 d: 



tAt of the correct volume, f 7ra 2 &. 



'4 fa 

— by Simpson's First Eule, using values 
l x 



3. Find — and 
Ji x 

of the integrand for x— 1, f , f- , etc., and compare the results with 

1.09861 = log3 and 1.38629 = log 4. 



220. Simpson's Second Rule. — Using the same assumption 
for f(x), as in the First Eule, but dividing the range into three 
equal parts, taking the origin half-way between the ends of the 
range, and having given the values of /( — ?), f( — f), /(f), 
and f(§), where K — b — a, we can prove by a precisely similar 
discussion that 



I. 



f(*)^=f[/(-#)+3/(-f)+3/(?) + /(f)], 

"2 

or that 

Cb b — a 

j a f(x)dx=-g- (y 1 +3y 2 +3y s +y i ), 



238 The Calculus. 

the y's being the values of f(x) at the beginning, the points of 
trisection, and the end of the interval. This method may also 
be applied if we take 3n equal divisions of the range, giving 

£ f(x)dx = ^( yi + Sy 2 + 3 yti + 2yt 

+ 3y B + 3y« + 2y 7 + • • • • +y 8 »+i)> 

where the y's are the values of f(x) corresponding to the points 
of division. 

These are briefly written: 

»/(z)&= »;° [1,3,3,1], 



or 



- 1+3+3+2 . a . . +1 c 1 . 3 > 3 > 2 > — ' y- 

They are known as Simpsons Second Rule. 

221. Employment of the assumption that f(x) is expressed by 
a polynomial of degree n gives, for various values of n, other 
approximations; e. g., 

n=4=: £ f(x)dx=^-[7, 32, 12, 32, 7]. 

n=6: P /(^)^=^#[41, 216, 27, 272, 27, 216, 41]; 

a slight change in the ratios of these coefficients gives 

h ^ [42, 210, 42, 252, 42, 210, 42], 
or 

^ [1, 5, 1, 6, 1, 5, 1], 
a very valuable rule, known as Weddles' Rule. 



Series. 239 



222. Examples. 

1. Prove that if c — b — b — a, so that a, b, and c are equally 
spaced values of x, and f(a)=y 1 , f(b) = y 2 , f{c)—y^ then 

/(a;)- dx=-j^(5y 1 + 8y 2 -y 3 ). 

This is known as " the five-eight rule." 

2. Prove in detail Simpson's Second Rule. 

3. Find the volume of the larger segment cut by the plane 

x = -jr from the sphere formed by revolving x 2 + y 2 = a 2 about the 

#-axis. (Use Simpson's Second Rule.) -Ans. f ira 3 . 

dx 



4. Find the value of 



by Simpson's First Eule, 






o 1+x 2 

making ten divisions of the range, and compare your result with 
the value of tt = 3.14159265. 

IT 

5. To find 46 2 f 2 cos 2 <£ d<f> (example a Art 193 and 
Jo VI— i sm 2 r ' 

example 3, Art. 206), compute the function at intervals of 7J° 
and apply both the first and the second of Simpson's Rules. In 
computing the values of the integrand, use an auxiliary angle 
such that sin 2 <f> = 2 sin 2 6 ; then the values of the function can 
be computed from the form : 

<j> log sin 2 log cos 

9.84949-10 
9 log sin log sec 

f{4>) log 



223. Evaluation of-~ Problem: To determine the value 

7X1 X^) 

to assign to a fraction — >-' when x = a if u (a) =0 and v(a)=0 } 

in order to complete the definition of an otherwise continuous 
function. (See Algebra, Art. 45.) 



240 



The Calculus. 



We have by Taylor's Series : 

u(x) =u\_a+ (x — a)] =u(a) + (x-a)u'(a) 



+^# «"(*)+ — , 



v(x) =v[a+ (x — a)] =v(a) + (x— a)v'(a) 
Since u(a) = and v(a) = 0, 



+ I*^LV'(«)+.... 



w(:r) 



-y (#)_ 



(a;-a)tt'(a) + ^r#^- w"(a) + 



(x-a)v'(a)+ ( X , 2 a)2 i/'(a)+. . . . 



tt '( a ) + T9~ w "W+ • * • 



v'(a) + ^v"(a)+ 



Thus we have, if u(a)=Q and v(a)=0, 



u'{a) 
~¥{a) ' 



If the fraction 



u(x) 

V 0*0. 

u'(x) 

~V(x) 



u'(x) 
, resulting from differentiation of the 



u( X i 

terms of -7— r , is indeterminate when x—a (that is, if v! (a) = 
v(x) 

and v'(a) = 0), the same method gives 



u"(x) 
V(s). 



as its value 



and so on. If differentiation of this sort is kept up after the 
fraction becomes determinate, the results will be correct only by 
accident. 

It is often more convenient, especially when a=0, to write 
the actual developments of u(x) and v(x) ; for if these develop- 
ments are known, all the differentiation is avoided. Only the 
first non-vanishing term of the development need be written. 



Series. 



241 



For example: 
sinz 



(a) 



Llog(l + aOJ 



x=0 



cos a; 



1 + x 



- T -i; 



x=0 



or 



smx 



log(l + a;) :c + 



(b) 



0-*sin0 



l + cos0 + cos20Je = £ 



[-si 1 



— ; cos 



1 

-l 



sin 0-2 sin 20 
-1 



Or, putting 6=\—<j>> the fraction is 



<£_£ cos <£ 



l + sin</> — cos 2</>J^=o 



I-+- 


.<■• 


2 


+) 


! + </>- 


-1 + 


(2«>) 2 
2 


+ 



= -1. 



4>=o 



224. Evaluation of — . — When ^) x {- defines a continuous 
co v{x) 

function except for x—a, and u(a) = oo and v(a) = co, the 

proper value to be given to the fraction in order to complete the 

definition may be found by writing 



u(x) _ y(x) 
v (x) 1 

u(x) 







which has the form —when x — a, and so may be treated by thi 

methods just given. 
For example, 



sec 



sec 50 
17 



00 



cos 50 



cos 



5 sin 50 
sin 



= 5. 



242 



The Calculus. 



225. The following theorem facilitates the evaluation in many 

cases : 

If u(a) = oo and v(a) = oo, 



'u(x) 



v(x) 



oo 

00 



u'{x) 



v'{x) 



In the following proof, the x's are omitted for brevity : 







" 1 1 




~ — dv 




~u~ 

V 


a 


V 

T 


= 


V 2 


= 


— du 






. u _ 


a 


L u2 


a 



dv 
du 



Hence 

or 
u(x) 



du 



dv 



v(x) 



du(x) 



dv(x) 



u 

V 


a 


U 

. v 


a 




dv 
du 


a 


[-1 ' 






u f (x)dx 
v'(x)dx_ 


x=a 


= 


\u'(x) 


x=a 



This theorem should be made nse of only when it gives forms 
simpler to invert than the original ones ; reduction to — is 

necessary at some time in handling — fractions, for without it, 



every derived expression will remain 



For example, 



log(s-l) 

[>g(z 2 -l)J 



x=l 



00 

00 



_1_ 

X-l 

2x 
x 2 -l 



x + ~\ 

2x 



= 1. 



x=\ 



Jar=l 



226. Any factor of an indeterminate form may be evaluated 
separately, unless the factor is zero or infinite. 



Series. 



243 



For example : 



cos 20 



l-tan0 



1 — COS X 

_x\og(l + x) _ 



(cos 2 fl-sin 2 fl)cosfl" 
cos — sin 



= (cos 6+ sin 0) 



t 1 



'=T 



cos — sin 0)cos 
(cos — sin 0) J 0=J 



a:=0 



1 — COS X 



x 2 log(l+£) 



log(l + ar). 



x=0 



x=0 



_ 1 

— 2" 



1 
1+Z 



1. 



= h 



x=0 



227. The Forms Ox oo, 1°°, 0°, and oo .— Other types of inde- 
terminate forms may be treated by first reducing them to the 

form -pr- or — , as follows : 
oo 

Oxoo: To evaluate [<f>(x)> 0(x)] x=a if <j> (a) =0, 0(a) = oo ; 

write 



[+(*)•»(*) Ua' 



or 



[0(*)]-\ 



x=a 

00 



1*: To evaluate [[<£('#) I'^l^a if <j>(x)=l, 9(x) = oo, 
evaluate 

[log[*(*)]«W] <M = [«(*)• log +(*)]_„= oo • 0. 
If the result is b, 

ll<l>(x)] m ] x=a = e». 

0°, oo°: To evaluate [[<f>(x)'] 9 ^] x=a if <£(*)= or oo, 
0(a) =0, evaluate 

[tog[*(*)]^ =[«(*)-l^+(*)]^0(:i:oo). 

If the result is &, 

[[+(*) ]**]«=•». 



244 



The Calculus 



228. Examples. 

Determine the following: 
tan x— x 



x — sin x 
1-x 



log X 

X' 



= 2. 

= — 1. 



z=l 



.1 — cos mx 
a x — b x 



1. 
2. 

3. 
4. 
9. 

10. 

n sinaj-sin nx 

11. [a tan x — £ sec a 



5. 



!* — <?-* — 2a ;] 
a — tana 



= -1 



6. [(1-*)"-].= V 
tana 



x 



cc=0 



ce=0 

= log 



m" 
a 



. lo S x 
/ tan a y/* 2 " 



=0. 



= 6*. 



m sm a; — sm ma 
a: (cos x — cos ma) 
tan nx—n tana 



m 



cc=0 3 



x=0 



X=-^ 



= 2. 
= -1. 



. (Use series.) 
(Use series.) 



12. [sec0-tan0] fl= ! = O. 

13. [sin a tan x ] *= = [tan a sin x ] x=0 = 1. 

14. [(sina)^ 2 -]^=yy/i. 



e 

1 
15. 



(cos mx) n,x * 

J a;=o V e T 

16. Trace the curve y—x 2 logx . 

17. Trace the curve y=x~ 3 logx . 



CHAPTEE IX. 

Mean Values. 

229. The sum of a set of quantities divided by their number is 
the average or mean of the quantities. That is, if a x , a 2 , a 3 , . . . . , 
On are n quantities, their mean is 

— (a x + a 2 + a 3 + .... + a„). 

Consider the following problem: Two straight roads, AB 

and AC, make an angle of 30° ; 
from A to B is 1 mile, and along 
AB telegraph poles are set 110 
feet apart; to find the average dis- 
tance of these poles from AC. 

Let any pole be P, its distance 
from AC be PP', and let AP=x; 




Fig. 78. 



then PP'= ±. 



There are 48 dis- 



tances, 55 feet, 110 feet, 165 feet. 
220 feet, etc. ; and their mean is 

A-X55X (1 + 2 + 3 + 4+ .... +48) =13471 ft. 

Now suppose that instead of 48 posts 110 feet apart there 



were n posts 
would be 

"5280 



5280 



feet apart; their mean distance from AC 



2n 



(1 + 2 + 3+ ....+») 



= 1320 



(■+-> 



If the mile of road AB is bordered by a fence having pickets 
2 inches apart, the mean distance of all the pickets from AC is 
1320 ft. \ in. 



246 The Calculus. 

It is natural to call the mean distance of the roadside from 
AC, 



1320+ \ 



1320 ft. 



230. This last result might have been got as follows : Using 
x as before, suppose points uniformly distributed along AB, h 
to the foot; and suppose AB divided into parts each dx feet long. 
If PQ is any one of these parts, it contains kdx points, each of 

which is approximately — feet from AC; the sum of the dis- 

a 

tances from A C of all the points in PQ is approximately ^rkdx 

feet, and the number of these distances is k. The sum of the 
distances from AC of all the points of AB is approximately 

<c=5280 

2\ -9- kdx feet, 

#:=0 

and the number of these distances is approximately 

£=5280 

? t kdx. 

cc=0 

The mean distance in feet is approximately 



and is exactly 



7, ~2 kdx /M-K~dx 

%kdx %dx 

580 x_ dx (5280) 2 



5280 dx~ 5280 



1320. 



jo 
We might, with the same result, have supposed each of the 

distances for points in the element to be -|- feet, and have called 



Meax Values. 



247 



the number of points in the element equal rather than propor- 
tional to the length of the element; when the idea at the basis 
of the process is clear, it is well to abbreviate in this way. 

231. Illustrative Examples. — Eequired the mean distance 
from the base of points on a semicircumference of radius a: 
(1) regarded as the limit of such a mean distance for points 
distributed uniformly along the arc, (2) regarded as the limit 
of such a mean distance for points whose projections are uni- 
formly distributed along the base. 

The arc in (1) or the base in (2) is called the region of uni- 
form distribution. 

In (1), radii to the points bound equal angular divisions; call 

one of the divisions dO; the equal 
divisions of the arc, the region of 
uniform distribution, are each a dO 
long. Let points be uniformly dis- 
tributed along the arc so that there 
are lcad$ in each element; the dis- 
tance for each is approximately 
a sin 6, the sum of the distances for all the points in the element 
is a sin 6-kadd approximately; the exact value of the mean, 

fa sin • k • adO 
o . 




Fig. 79. 



M 



= a -L° 



sin d6 



2a 



= a 



ka • dO 



dO 



Jo jo 

In (2), divide the region of uniform distribution, the base, 

into parts e&ch.=dx J each containing 
kdx points, for each of which the 
distance is approximately Vaf — x 2 : 
Sum of distances for an ele- 
ment = k V^ — ^dx. 

Sum of distances for all the ele- 




Fig. 80. 



ment = 2k \/a 2 — x 2 • dx. 



248 The Calculus. 



a 



Number of distances taken in all the elements = 2& dx. 

—a 

Mean value = — - — ~~ x — - approximately. 

^ax 

Limit, or mean distance required 

cos 2 e d6 



[° Va 2 -x 2 -dx a ) 
— ]—a : 



ax 

—a 



The mean distance in (1) is about 0.6366a; in (2), is about 
.7854a. It is clear that the distances of points near the base, 
where the arc is nearly vertical, have counted much less heavily 
in (2) than in (1). 

Note that a mean value is not defined unless the region of 
uniform distribution is given. 

232. In order to find the mean value of a given function for 
a given region of uniform distribution, divide this region into 
elements throughout each of which the function may be assumed 
to have the same value; find the integral of the product of the 
function by the element throughout the region, and the integral 
of the element itself with the same limits. The mean value will 
be the quotient of the first integral divided by the second. 

The region of uniform distribution may be, as in the cases 
cited, a length, or it may be an area or a volume. Other func- 
tions than functions of position are treated in the same way; 
for instance, the average distance fallen by a body in a given 
time may be computed as the limit of the average of the dis- 
tances fallen as it reaches points distributed at uniform distances 
from top to bottom of its path, or as the limit of the average of 
the distances fallen at instants of time distributed at uniform 
intervals from start to finish of its fall. 



Mean Values. 249 

Lines passing through a fixed point and uniformly distributed 
about it (so as to divide the space around about into equal plane 
or solid angles), if they lie in one plane, mark points uniformly 
distributed around any circumference centered at the fixed 
point; and if not so confined, mark points uniformly distributed 
over the surface of any sphere centered at the fixed point. 

233. Examples. 

1. Find the average distance of points of a circle from the 
center, and the average distance of points of a sphere from the 
center, the regions of uniform distribution being the area and 
volume respectively. Ans. \a and fa. 

2. Show that the mean distance of points on the circumfer- 
ence of a circle from a fixed point on the circumference is — . 

7T ' 

region of uniform distribution the circumference. 

3. Show that the mean distance of points of a circle from a 

fixed point of the circumference is -q^-, points uniformly dense 

over the surface. 

4. Find the average distance of points of the surface of a 

hemisphere from the base, the region of uniform distribution 

being (a) the hemispherical surface, (b) the base. 

A a 2a 

Ans. -, -g-. 

5. Find the mean distance of points uniformly distributed 

over the surface of a sphere from a fixed point on the surface. 

a 4a 

Ans. -g_. 

6. Find the mean length of chords drawn from a fixed point 
on the circumference of a circle and uniformly distributed about 

the point. Ans. — . 

7. Find the mean distance of points uniformly distributed 
throughout a sphere from a fixed point on the surface. 

a 6a 

Ans. - ¥ -. 
5 



250 The Calculus. 

8. Find the mean length of chords drawn from a fixed point 
on the surface of a sphere and uniformly distributed about the 
point. Ans. a. 

9. Examples 2 and 6 have the same result; examples 5 and 
8, different results. Why? 

10. Show that the average distance from the base of points 
uniformly distributed through the volume of a hemisphere is f 
of the radius. 

11. Show that the average latitude of points uniformly dis- 
tributed over the surface of the northern hemisphere of the earth 
is about 32° 42'. 

12. Show that the mean distance of points uniformly dis- 
tributed along the perimeter of a square from one corner is 0.8239 
of the side. 

13. Show that the mean distance from one corner of a square 
of points uniformly distributed over the area is 0.7652 of the 
side. 

14. A line is divided at random into two parts. Find the 
mean of the product of the segments so formed, the points of 

a 2 

division being uniformly distributed. Ans. -^-. 

15. Show that for the ellipse x=aeo§<j>, y — bsmcj), if the 
length of a quadrant is Q, the mean radius of curvature for 
points uniformly distributed along the arc is 

(3a 4 + 2a 2 & 2 + 36 4 ). 



IGabQ 



MECHANICAL APPLICATIONS 



CHAPTER X. 
Kinematics. 

234. Displacement— Kinematics treats of the relation between 
change of position and the time in which change of position takes 
place. A change of position from a point A to a point B is meas- 
ured by what is called the displacement AB, which is determined 
by the distance AB and the direction of B from A. Just as the 
distance from A to B is defined as the length of the straight line 
joining A and B, and is, therefore, independent of whatever path 
may actually be traced by a point moving from A to B. so the dis- 
placement AB has nothing to do with the path of motion, but is 
wholly determined by the relative positions of A and B. The 
direction of a displacement is as important as its numerical mag- 
nitude or distance. 

235. Two displacements are equal if they produce the same 
change of position; that is, if they have the same distance and 
direction. The sum of two displacements is the single displace- 
ment which causes the same change of position as the two dis- 
placements combined. It is found as follows : Let the given dis- 
placements be p and q, Fig. 80a ; let the point A be given the dis- 
placement p, which carries it to B, determined by drawing AB of 
the length and direction of p; let the displacement q be given to 
the point B, so that it is carried to C(BC — and || to q and in 
the same direction) . Then A is carried by two successive displace- 
ments which result in the new position C. This change of position 
is measured by the single displacement AC, or r, which is thus the 
sum of the displacements p and q. 

p + q — r. 



252 



The Calculus. 



If the displacement q is made first, and followed by the dis- 
placement p, A will be carried to the same final position 0, as is 
evident from the parallelogram ABCD of Fig. 80a. The sum of 
two displacements is, therefore, independent of the order in which 
they are combined. 

Indeed, the sum of two displacements is the same even if the 
two take place wholly or partly in the same interval of time. For 
instance, suppose a man to be rowing a boat in a river, and suppose 
q in Fig. 80a to be the displacement that would result from his 
rowing if there were no current, and p to be the displacement that 
would be caused by the current if he did no rowing ; then r is the 
actual displacement due to the combined causes. 




w 



Fig. 80a. 



Then, in order to construct graphically the sum of two displace- 
ments, choose some convenient scale of distances, and draw a 
triangle ABC, giving AB and BO the directions of the given dis- 
placements, and making their lengths correspond to the distances 
of the displacements; the direction of the required sum will be 
from A to 0, and its distance, or numerical magnitude, will be 
represented, to the chosen scale, by the length of AG. 

The sum r, of two displacements p and q, is ordinarily called 
the resultant of the displacements, and the displacements p and q 
are called components of r. The addition of displacements is 
called composition; thus, the displacements p and q are said to be 
compounded into the resultant r. 

The angle from one displacement to another is the angle through 



Kinematics. 253 

which the first displacement must be turned in order to make it 
point in the same direction as the second; thus in Fig. 80a the 
angle from p to q is <f> = 180 ° — B. 

The distance and direction of the resultant (or sum) of two 
component displacements can be computed by the Law of Cosines 
and the Law of Sines of Trigonometry from the formulas 

r 2 — p 2 + q 2 + 2pq cos $=p 2 + q 2 — 2pq cos B, 

. a sin <£ 

i r 

236. If two displacements involve the same distance, and are 
opposite in direction (</> = 186 ), their combined effect is to cause 




Fig. 81. 

no change of position ; that is, their sum or resultant is zero. In 
this case, each displacement is the negative of the other. 

If we are given the resultant of two displacements and one of 
the component displacements and are required to find the other 
component — in other words, if we are required to subtract a dis- 
placement, we can do so by adding its negative. 

For instance, if we are to determine the difference p — qoi the 
displacements p and q, in Fig. 81, we have, from the right-hand 
triangle, p+ ( — q) — d, and evidently d is the required difference, 
since, from the left-hand triangle, q + d — p. 

237. If we are given the resultant of two displacements and the 
two directions of the components, we can determine the com- 
ponents as follows: Let r (Fig. 82) be the given displacement, 
and let the required directions of its components be those of the 



254 The Calculus. 

lines a and b. Draw to convenient scale and in the proper direc- 




Fig. 82. 

tion a line to represent r, and from its ends draw parallels to a 
and b. In either of the triangles thus formed, we have 'two dis- 
placements, p and q, whose sum is r, as required. The lengths of 
p and q can be computed by the Law of Sines, since 

sin (<£ — 0) n , sin 

p=r ^— ; — '- and q = r— — = . 

r sm<£ 2 sm<£ 

In this case, r is said to be resolved along the given directions 
a and b into the two components p and q. When the given direc- 
tions are perpendicular, (</> = 90°), the resulting components, 
p—r cos 6 and q = r sin 0, are called resolved parts. 

238. The most usual way of treating displacements, especially 
in problems that are at all complicated, is 
to resolve each displacement along the di- 
rections of a pair of rectangular coordinate 
axes, as in Fig. 83 ; the resolved parts are 
then evidently : x—r cos 0, y—r sin 8; and 
the resultant is determined from its com- 
ponents by: r 2 = x 2 + y 2 , 0=tan -1 $- — 




sin -1 ^- — cos - 



Two such resolved parts, referred to a given 



Kinematics. 255 

pair of axes, determine the resultant displacement completely. In 
fact, the distance and direction which define the displacement are 
merely the polar coordinates of the point to which the origin is 
carried by the displacement ; the resolved parts, on the other hand, 
are the rectangular coordinates of the same point. 

The resultant of any number of displacements is found by 
adding (or compounding) some two, adding the sum of these to a 
third, and so on, until each of the given displacements has been 
used. 

It is possible to compute the distance and direction of the re- 
sultant of a number of displacements from the trigonometric rela- 
tions already given, but this is by no means the easiest way. It is 
much simpler to choose a convenient pair of rectangular axes, 
resolve each of the given displacements along the two axes, combine 
separately the ^-components and the ^-components of all the dis- 
placements, and finally determine the resultant of the two per- 
pendicular displacements thus found. 

For instance, let it be required to find the sum or resultant of 

the following displacements: p x 2 mis. N. 20° E., p 2 2J- mis. N. 

135° W., p 3 3 mis. N". 15° E., p 4 1J mis. N". 20° W., p 5 45 mis. W. 

40° S. Taking the positive direction of the z-axis due east, we 

have 



r. 


0. x=r cos 9 




y=r sin.6. 




^=2 


1 = 7O° 3-! = 0.684 




y x = 1.879 




r 2 = 2.5 


2 = 125° x 2 = 


- 1.434 


ft = 2.048 




r 3 = 3 


3 = 75° x 3 = 0.776 




2/3 = 2.897 




k=1.5 


4 = 11O° x± = 


- 0.513 


ft = 1.410 




r 5 = 45 


6 5 =-l±0°x 5 = 


-34.47 


y 5 = 


-28.93 




1.460 


-36.42 
+ 1.46 


8.234 


+ 8.23 




X=%x= 


= -34.96 


Y=Sy= 


: -20.70 



256 The Calculus. 



A o4.yb 



R=VX 2 + Y 2 =X sec ® = Y esc © = 40.63. 

The combined effect of the five displacements is thus the same as 
that of a single displacement of 40.63 mis. W. 30° 28' S. 

239. Examples. 

1. A point undergoes the following displacements: 40 ft. 
N. 60° E.; 50 ft. S.; and 60 ft. 1ST. 60° W.; find the resultant 
displacement. Ans, 10 V3 ft. W. 

2. Show that two component displacements represented by two 
chords of a circle drawn from any point P and at right angles, are 
equivalent to a single displacement represented by a diameter of 
the circle. 

3. A ship makes 40 miles S. 30° E., 60 miles S. 60° W., and 50 
miles N". 30° W. ; find the resultant displacement. 

Ans. 60.83 miles N. 159° 28' W. 

4. A particle suffers five successive displacements of magnitudes 
a, 2a, 3a, 4a and 5a, parallel to the sides of a regular hexagon 
taken in succession ; what is the resultant displacement ? 

Ans. 6a, making 180° with the first. 

5. A steamer is carried by her propeller 12 miles N., and by the 
wind 3 miles S. 15° E., finds that her displacement is 15 miles 
NE. ; find the displacement due to a current, which is unknown. 

Ans. 9.94 miles N. 81° 18' E. 

6. A carriage wheel, 16 inches radius, rolls along a horizontal 
road; find the displacement of a point originally in contact with 
the road after the wheel has made a quarter revolution. 

Ans. (approx.) 18.4 in., at angle of 60° 17' with the horizontal. 

Speed, Velocity and Acceleeation. 

240. Speed. — The speed of a point moving in a straight line 
was defined in Arts. 7-10 as follows : 

If a point moves As feet in Atf seconds, v— — ^- is its mean 



Kinematics. 257 

speed. If this mean speed is constant throughout the motion, the 

As 
point moves with the uniform speed, v = — . 

If the mean speed is variable, the actual speed of the point at 
any instant is the value at that instant of the time-derivative of 

the distance traversed, i. e., v — —77- . 

These are also the definitions of uniform speed, mean speed and 
speed for motion of any kind. 

It should be observed that speed is in a^^ case a distance; in the 
case of uniform speed it is the distance actually traversed in each 
second ; the mean speed is a uniform speed, and a variable speed 
is the limit of a mean speed. 

The speed of a moving point is, therefore, represented graphic- 
ally by a length corresponding to the distance traversed in one 
second. 

For instance, if a point moves in 5 seconds over the quadrant of 

a circle of 10 feet radius, its mean speed is — '-= — = 3.1416 f/s, 

o 

and would be represented by a length corresponding (in accord- 
ance with a chosen scale) to 3.1416 feet. 

241. Velocity. — The definition of velocity is similar to the defi- 
nition of speed, but with the important difference that the dis- 
placement given to the moving point takes the place of the distance 
traversed by the point. 

If the velocity is constant, the same displacement must be given 
to the moving point in any two equal intervals of time ; that is, the 
point must move the same distance and in the same direction; in 
other words, it must move with uniform speed in a straight 
line. Motion with constant velocity is, therefore, uniform recti- 
linear motion. 

If a point moves in a straight line with variable speed, its mean 
velocity during any interval after a given instant, and its actual 
18 



258 



The Calculus. 



velocity at the given instant, are evidently determined in magni- 
tude by the corresponding speeds and in direction by the direction 
of the motion. 

The velocity of a point moving in any way is represented 
graphically by a displacement corresponding to the displacement 
given to the point in 1 second. In rectilinear motion, the mean 
velocity and the velocity are represented by displacements whose 
distances are the lengths representing the corresponding speeds, 
and whose directions are the direction of the motion. 

242. The velocity of curvilinear motion is always variable, for 
whether the speed changes or not, the direction must; conse- 
quently, although the magnitude of the actual velocity at any in- 
stant is the speed at that instant, the magnitude of a mean velocity 
is not the corresponding mean speed. 

Consider, for instance, the example of Art. 240, and suppose, for 
convenience, that the point moves over the quadrant of the circle 
of 10 feet radius in 5 seconds at uniform speed. 

Then, in moving over the quadrant AB (Fig. 84), the point 
covers 15.708 feet in 5 seconds at the 
uniform speed (which is also its mean 
speed) of 3.1416 f/s. This motion, how- 
ever, gives it a displacement AB of 
14.1428 feet in the direction from A to B. 
Its mean velocity is, therefore, 2.8286 f/s 
in a direction making an angle of 135° 
with OA. The mean speed is represented 
by a length corresponding to 3.1416 feet 
drawn regardless of direction; the mean 
velocity by a distance corresponding to 2.8286 feet at the angle 
135° with oA, as AM. 

To find the actual velocity at A, consider the motion from A to 
P, calling the angle AOP = A0. Then the arc AP = As = lOA0, 

10 A0 




and as As = 3.1416A^ At= 



3.1416 



is the number of seconds 



Kinematics. 



259 



A0 



required to reach P. The displacement AP is 20 sin — in mag- 
nitude, and makes the angle (90°— -s— ) with OA The corre- 



sponding mean velocity is a displacement, 20 sin 



A0 



10 A0 



(smjf \ 



3.1416 



3.1416 I ■ — — I in magnitude and makes the angle (90° 



A0 

9 



with OA. Since 



sin a: 

x 



1, the limit approached by this 



mean velocity as At and hence A0 becomes zero, is the displacement 
3.1416 at 90° with OA, represented by AV 1 in the figure. The 
velocity at A thus has for its magnitude the speed at A, and for its 
direction the direction of the motion at A. In the same way, the 
velocity at any point of the path will be seen to have the speed of 
the motion for its magnitude and the direction of the motion for 
its direction. If the motion continues around the circle, the 
velocities at B, C, D are represented by the displacements BV 2 
CV S , DV A , from which it appears that the change in velocity be- 
tween opposite points of the circle causes a reversal of direction, 
and amounts, therefore, to 6.2832 f/s. 

243. There is no material difference in the most general case of 

curvilinear motion. Let a point move in any way along a curve 

AB; to determine its velocity when, after t 

seconds, it reaches the point P. Consider its 

motion from P to P' ; call the arc PP' — As. 

the chord PP' — Ac and the time taken to move 

from P to P' — At. Then the mean velocity is 

Ac 
——-in the direction of the chord PP' ; and 

the mean speed is -^- ; the limit of the mag- 
nitude of the mean velocitv when A£ = is Fig. 85. 




260 The Calculus. 



—tt- = —77- , the speed at P, and the limiting direction of 

^ J At=0 at 

the velocity is that of the tangent at P or of the path of motion 
at P. 

In any case of curvilinear motion, then, the velocity at any 
point of the path has for its magnitude the speed at that point, 
and for its direction the direction of motion. 

244. Since a velocity is a displacement (the displacement that 
would take place if the moving point preserved its motion un- 
changed for 1 second), the addition and subtraction of two veloci- 
ties, or the composition and resolution of any number, has already 
been discussed in the treatment of displacements. 

For instance, in the discussion just preceding, we might have 

referred the motion of the point P to rectangular coordinates. 

Then the displacement PP' might have been determined by its 

two components, found by resolving it along the axes. If P has 

the coordinates (x, y) arid P' the coordinates (x + Ax, y + Ay), 

the components of the displacement PP' are Ax and Ay in the 

Ax 
directions of the axes; the mean component velocities are — and 

— — in these directions and the velocity at P has for its com- 
At J 

dx dv 
ponents in the directions of the axes the limiting values -, and -j? . 



The resultant velocity is, therefore, 

•(-*; + m 

in the direction making the angle 

tan Kdt^dl 

ds 
with the #-axis, or is -rr in magnitude, and 

^ 0(y makes the angle tan -1 -J^ with the #-axis, and 

Fig. 86. ° dx 




Kinematics. 261 

so is again seen to have the magnitude of the speed and the direc- 
tion of the motion. 

245. Examples. 

1. A ship sailing north at a speed of 8 m/h is carried east by a 
current of 4 m/h ; find the resultant velocity. 

Ans. 8.94 m/h, K 26° 34' E. 

2. A point moves N". 30° E. 60 feet in 10 seconds, then west 30 
feet in 20 seconds; find the mean speed and the mean velocity. 

Ans. Mean speed 3 f/s, mean velocity V3 f/s N. 

3. What will be the apparent velocity of rain drops falling 
vertically 20 f/s to a person in a train having a speed of 30 m/h ? 

Ans. 43.33 f/s, at 24° 27' with the horizontal. 

4. A ship steaming 8 m/h due east has an apparent north wind 
of 6 m/h ; what is the velocity of the wind ? 

Ans. 10 m/h N. 53° 8' W. (nearly). 

5. If the mean speed of the earth in its path around the sun is 
18.6 m/s, and the speed of light is 186,000 m/s, what is the ap- 
parent angular displacement of the sun, due to the combined 
motion of the two ? 

Ans. 20" 6 in a direction opposite to the earth's motion. 

6. A ship heading east at 10 knots an hour, makes 10 knots an 
hour NE. due to a current; find its velocity. 

Ans. 7.65 knots an hour N. 22^° W. 

7. Find the velocity of a point on the rim of a wheel, radius a, 
rolling along a straight line. 

Ans. v=4a sin -=j- , in a direction 90° —6 to the given line. 

(0 = i the angle of revolution measured from the lowest point.) 

8. Find the velocity of the point of problem 7, when the center 
of the wheel moves uniformly at a speed of 60 m/h ; &=4 feet and 

<f>=26= ~ and ir. 

Ans. v = 88V2 f/s for <f>= -^-, v = 176 f/s for <f> = ir. 



262 The Calculus. 

9. A sailing ship heads NE. and makes a speed of 15 m/h, at 
the same time the wind carries her 2 m/h SE., and a current 
4 m/h N". 30° W. ; what is her velocity over the ground ? 

Ans. 16.14 m/h K 38° 23' E. (approx.). 

246. Acceleration. — Acceleration was defined in Art. 75 foi 

rectilinear motion as the time-rate of speed: v = ~ , a= ~ = 

at ' at 

This special case of acceleration is the distance a second 



dt 2 

that the speed increases in each second, or, if the acceleration is 
variable, is the limit of such an increase. 

Acceleration is defined in general as the time-rate of increase 
of velocity, and so is the displacement a second by which the 
velocity increases in each second, or the limit of such an increase. 
As an acceleration is thus a displacement, the composition, resolu- 
tion, etc., of accelerations is merely another special case of the 
performance of these operations on displacements. 

247. Acceleration in Rectilinear Motion. — In the case of recti- 
linear motion with uniform speed, the acceleration is zero, since 
there is no change of velocity. For rectilinear motion with 
variable speed, the change in velocity in any interval of time is a 
velocity in or opposite to the direction of motion — for if the in- 
crease in velocity were in any different direction, its addition 
would change the direction of motion. 

Hence, for rectilinear motion, the mean change in velocity dur- 
ing the interval, or the mean acceleration, is in the direction of 
motion for any interval of time, and its limit, the acceleration, is 
also in the direction of motion. 

It is, however, only in the case of rectilinear motion, that is, 
motion in which the direction of the velocity does not change, that 
the acceleration is in the direction of motion. 






Kinematics. 263 

248. Acceleration in Uniform Circular Motion. — Consider, for 
instance, the case next in simplicity to rectilinear motion : uni- 
form circular motion, in which a point moves along the circumfer- 
ence of a circle at a constant speed. Let the radius of the circle be 
a feet, and the constant speed z f/s. The 
velocity at any point in the path has for its 
magnitude the speed, z f/s, and for its di- 
rection the direction of the motion, and so 
may be represented by a tangent to the path 
at the point in question, marked in the direc- 
tion of the motion, of length to represent z 
feet. Let PV represent in this way the 
velocity v at a point P, and P'V the velocity Fig~~87 

v + Av at a point P' , the arc PP' being As 
feet and subtending the central angle A$. Construct the displace- 
ment P'R to represent At' = v + Av + ( — v ) . Then P'R = Av is the 
change that occurs in the velocity of the moving point as it goes 
from P to P' . Dividing Av by the number of seconds the motion 
takes, we shall have the mean acceleration in the corresponding 
interval. The distance PP' = As=aA0 is covered uniformly at 

z f/s, taking seconds. The mean acceleration is, therefore, 

In the triangle P'V'R each of the sides P'V and RY' 




aAO 

represents z feet, and the angle V — AS. Hence the side P'R rep- 
resents 2z sin 4r f eet - The an S le Rpr v ' = 90 ° ~ 4r • Thus A" 5 
magnitude of the mean acceleration is 



zAv 


Ad 
zX2zsm -o- 


z 2 


AS 

sm ~2~ 


aAd 


aAO 


a 


AS 

9 



264 The Calculus. 

and its direction is at the angle 90° — with the direction of 

motion or is at the angle -=- with the radius P'O (directed from 

P' to the center ) . In the limiting case of the acceleration at P, 
(A0=O) we therefore have: The acceleration at any point of 

z 2 
the path in uniform circular motion is - — in magnitude, and is 

directed toward the center of the circle ; z being the speed of the 
motion, a the radius of the circle. 

249. It should be observed that in all the preceding discussions 
we have used foot and second as units of length and time; the 
statements would have been more general if we had written " unit 
of length " and " unit of time " in all cases, but the definite units 
are simpler. Our results, of course, apply in the case of any units, 
but it is necessary that the units used in any particular discussion 
should be the same throughout. 

250. The treatment of uniform circular motion by rectangular 

components is instructive. Let the 

Y| point move as before in a circle of ra- 

^^| ^\ dius a ^ ee t a "k a uniform, speed of z f/s. 

/ /a— ^ -\p Eefer the motion to a pair of perpen- 

L ^ ^9^\\ dicular axes, OX and OY, drawn 

T °\ x M \ A X through the center of the circle, and, P 

\ J being any position of the moving point, 

\^!__^/ let (x, y) be its coordinates and the 

angle XOP = 0. Let us find the ac- 

FlG 88 celerations of M and N, the projections 

of P on OX and OY. x—aco^, 6, y = 

a sin 6, arc AP = a6; and since v.. ' =z, -jt = — , a constant. 

at 'at a 



Kinematics. 265 



Call — = — u — <°- 
a at 

dx . A dO dy n dO 

dt dt ' dt dt 

dx • n dv n 

—— = — a<D sm 6, —2— — (Jam COS 0, 

dt at 

d 2 x d^u 

—— — — aw 2 cos 6, -^ — — cud 2 sin 6. 

Then the velocity of motion is given by the components — am sin 
in the direction of the z-axis and aw cos 6 in the direction of the 



y-axis, or is aw V sin 2 0-fcos 2 d=aoi~z in the direction making 

, , aw cos ■ , am cos 6 , -> , , A * • m n\ 

tan" 1 = — j. ^sim 1 = tan" 1 ( —cot 0) =sm- 1 (cos 0) — 

— aw sin aco \ / \ / 

90° +0 with the cc-axis. This merely reproduces the given con- 
ditions. 

The acceleration has for its components — aw 2 cos in the direc- 
tion of the a:- axis, — aw 2 sin 6 in the direction of the ^-axis, and so 

a<D 2 sin 6 



is a<o 2 in magnitude and is directed at the angle tan -1 



— aw 2 cos 



sin- 1 m 2 = tarn 1 tan 6>= sim 1 ( - sin d) = 180° + 6 with the 



tc-axis. Since m= — and a<o 2 — , the results of the earlier dis- 

a a 

cussion are thus verified. 

251. In the general case of motion in a plane curve, the ac- 
celeration is determined as follows: Let a point moving in a 
plane curve (Fig. 89) be at P and At seconds later be at P'. Let 
the arc PP' = As, and let the velocities at P and P' be v and v + Lv, 
represented by the displacements PV and P'V . Construct the 
increment &v = P'R; then the mean acceleration in the interval 

P'R 

of At seconds has the direction of P'R and the magnitude — r— . 



266 



The Calculus. 




Fig. 89. 



It is convenient to treat the compo- 
nents of the acceleration in the direction 
of motion at P and in the perpendicular 
direction. Eesolving Av in these direc- 
tions, we have a mean component ac- 
celeration of magnitude — - in the di- 
rection of motion at P, and a mean com- 

P'M 



At 



ponent acceleration of magnitude 

in the direction of the interior normal at 
P. Let A<£ be the angle between the 
tangents at P and P' ; then the angle 
P'V'M = A<f>, and 



MR-P'V cosA<l>-RV'=(v + Av)cosA<f>-v 
= Av cos Acf> — 2v sin 2 -^ . 



The magnitude of the component of the mean acceleration in the 
direction of the tangent at P is thus 

A<£ 



MR Av AJL • A<£ 

— = —cos A+-V sin - 



sin 



i±* 



A<f> 

At 



The limit of this, when A* = 0, is ^ Xl-^xOxlX ^r = 4r, 
9 dt dt dt 

since, when A£ = 0, A<£ = 0. 

The component in the direction of motion at P of the ac- 

dv 
celeration at P is thus ^r > where v represents the magnitude of 

d s di> d s 

the velocity, or the speed, -rr , at P. This component, —^ — -^ , 

is commonly called the tangential acceleration or acceleration in 
the path of motion. 



Kinematics. 267 

Further, 

P'M - P' V sin A<£ = ( v + Av ) sin A<£. 

The magnitude of the component of the mean acceleration in the 
direction of the interior normal at P is thus 

P'M sinA<£ , Ay . A , sin A<£ A<£ Ay . A , 

The limit of this, when At = 0, is 

d<j> _ ds d<f> ^2 d<p v 2 



dt dt ds ds p } 

ds 
since, by Art. 80, -=-- is the radius of curvature, p. 

The component in the direction of the interior normal at P of 

v 2 
the acceleration at P is thus — , where v represents the magnitude 

ds 
of the velocity, or the speed, - , at P, and p is the radius of curva- 
ture of the path of motion at P. This component is commonly 
called the normal acceleration at P. The normal acceleration is 
always directed along the interior normal, hence the positive 
value must always be used for p in the formula. 

Finally, then, the tangential and normal components of the ac- 

dv d 2 s 
celeration of a point moving in a plane curve are a t = -rr = -p- in 

v 2 
the direction of motion, and a n = along the interior normal, 

ds 

v= -j- being the speed. 

Consequently, the total acceleration is 

,/ 2 , 2 _ // d 2 s \ a , v* 



268 



The Calculus. 



in magnitude, and is directed at the angle 




tan - 



= tan - 



d 2 s 



with the direction of mo- 



P 



dt* 



tion, on the same side of the tangent as the curve 

itself (Fig. 90). 

In the special case of rectilinear motion, we 

d 2 s 
have p = oo y a n — 0, so that at = —th- is the total 

acceleration. 
In the special case of uniform curvilinear motion, we have 



v constant, 



0, so that a n = — is the total accelera- 
P 



dv _ d 2 s 

tion. As a particular example of this case, we have the uniform 
circular motion already discussed in Art. 248. 

In the general case of curvilinear motion, the normal accelera- 
v 2 



tion, a n - 



, arises from the changing direction of the velocity; 



d 2 s 

—r^ , arises from the variation of 
dt 2 



the tangential acceleration, a* 

ds 

the speed, v = -vr , or magnitude of the velocity. To emphasize 

O/V 

this relation, a n is sometimes called the shunt, at the spurt of the 
acceleration. 



252. 



Examplt 



1. A point moves in a circle of 4 feet radius with a uni- 
form speed of 4 f/s ; show that the magnitude of the acceleration 
is 4 f/s 2 . 

2. Derive from the component velocities and accelerations the 
total acceleration in example 1, and show that it is directed 
towards the center (assume the path of mlotion a;=4eos0, 
y = 4:SmO). 

3. A point moving in a given direction with a speed of 600 f/s, 
one minute later is moving with the same speed in a direction of 



Kinematics. 269 

60° with the first; what is the total change in velocity and the 
mean acceleration? 

Ans. 600 f/s and 10 f/s 2 in a direction of 120° with the first. 

4. The earth's equatorial radius is 3962.8 miles; find the 
velocity and the acceleration of a point on the equator due to the 
earth's rotation on its axis. 

Ans. 1522 f/s and 0.111 f/s 2 . 

5. If the value of g on the equator is 32.1 f/s 2 , find what hori- 
zontal velocity a point must have in the plane of the equator to go 
around the earth. 

Ans. 27,438 f/s E. or 24,394 f/s W. 

6. The mean distance of the moon from the earth is 238,800 
miles, and its mean period of revolution about the earth 27^- days 
(approximately) ; show that its mean speed is 3354.6 f/s and its 
acceleration 0.0089 f/s 2 = 0.107 inch per second. 

7. Show that tangential and normal accelerations of a point on 
the rim of a wheel of radius a rolling along a straight, line are 

— -4- A-a r.ns H — 



a t = 4a sin 6 -^ + 4a cos I -=- J and a n = 4^a sin (-=-) , where 6 

is one-half the angle through which the wheel has turned. 

8. If the center of the wheel in example 7 moves at the uniform 
speed of 60 m/h, and a— 4 feet, show that the total acceleration 
has the constant magnitude 1936 f/s 2 and is directed toward the 
center of the wheel. 



CHAPTER XI. 

Forces. 

253. Words are often used as scientific terms in senses rather 
different from their ordinary meanings ; for instance, the distinc- 
tion between velocity and speed that we have been obliged to make 
for the sake of accuracy is foreign to every-day use. The term 
force, however, has in mechanics the meaning with which every- 
body is familiar. The forces most commonly in evidence are 
probably the pushes, pulls and twists exerted by one's own muscles 
and the pull or attraction of the earth, the force of gravity, which 
is exerted on all material bodies. The simplest force with which 
to experiment is the pull of a stretched coiled spring or rubber 
band, for if such a spring is kept stretched to a fixed length, it 
exerts a constant pull. The force of gravity is not so simple, be- 
cause it acts more strongly on some bodies than on others. 

If a material body of any sort is placed on a very smooth hori- 
zontal table and pulled with a constant force (by a spring, for 
instance), the pull will be practically the only force acting. Under 
these circumstances, the body will move with increasing velocity, 
but constant acceleration. If this body and another just like it are 
acted upon together by the same pull, the acceleration will be half 
as great, and a pull strong enough to move both bodies with the 
original acceleration will move one of them alone with twice the 
acceleration. Again, suppose we have a number of bodies all alike, 
made of steel, say, and a second set all alike among themselves, 
made of wood. Suppose the wooden bodies are large enough so 
that when ten of them together are acted upon by a certain pull 
they acquire the same acceleration that is given by this pull to three 
of the steel bodies. Then it will be found that the force required 



Forces. 271 

to give a certain acceleration to one wooden body will give -^o this 
acceleration to one steel body. 

As a result of experiments of this sort, though of course most 
elaborately and carefully made, certain facts and definitions have 
been established, on the basis of which the science of mechanics 
has been developed. 

254. First, it is found that any force, acting by itself on any 
material body, gives to the body an acceleration in the direction 
of the force. Also, if the magnitudes of different forces are de- 
fined to be proportional to the accelerations they give to the same 
body, it is found that the same relative magnitudes will thus be 
assigned to all forces, no matter what body is used in testing them. 

As a corollary of this law, it appears that if no force acts on a 
body, the acceleration of the body will be zero ; that is, the body 
will be at rest or else moving with uniform speed in a straight line. 

It is found that the size and material of a body affect the mag- 
nitude of the acceleration it receives from the action of a force,, 
and that if the masses of different bodies are defined as magni- 
tudes inversely proportional to the accelerations given to the bodies 
by the same force, the relative magnitudes thus assigned will be 
independent of the force used in the tests. 

255. Law of Motion. — All the preceding is summed up in the 
Law of Motion: 

If a force f, acting on a body of mass m, gives to it an accelera- 
tion a, the magnitude of f is proportional to the product of the 
magnitudes of m and a, and the direction of f is the direction of a. 

This relation of the magnitudes of f, m and a is expressed in 
the 'Equation of Motion: 

f — hma. 

256. Units. — The value that must be given to the constant k 
will of course depend upon the units chosen for acceleration, 
mass and force. It is desirable to make the equation of motion as 



272 The Calculus. 

simple as possible ; consequently the relation between the units is 
always made such that the unit force will give unit acceleration 
to a body of unit mass; then l = fcx 1 X 1, or 35?= 1, and the equa- 
tion of motion becomes 

f — ma. 

There are four systems of units in use; in the first place, the 
English Systems take 1 f/s (one foot a second each second) as 
the unit of acceleration, and the Metric Systems take 1 cm/s 2 . 
Further, in the Gravitational or Engineers Systems, a familiar 
force is chosen as the unit of force, so that the mass to which this 
force will give unit acceleration must be the unit of mass, while 
in the Absolute Systems, the mass of a well-known body is taken 
as the unit of mass, so that the force which will give it unit ac- 
celeration is the unit of force. The basis of all these systems is 
the fact, experimentally established, that the earth's attraction 
(the force of gravity), if unimpeded, will give the same accelera- 
tion to any two bodies in the same situation. The value of this 
acceleration is indicated by g; it varies for different situations, 
and is not even constant over the surface of the earth ; its surface- 
value, however, is never far from 32.2 f/s 2 or 981 cm/s 2 . 

The force with which gravity acts on a body (under certain 
standard conditions) is called the weight of the body; the weight 
of a piece of platinum kept in the Standards Office in London is 
called one pound, and the weight of a piece of platinum in the 
Palais des Archives in Paris is called one kilogram. 

I. In the English Gravitational System, the unit of force is the 
pound, the unit of acceleration is 1 f/s 2 , and the unit of mass is 
the mass to which a force of 1 pound would give an acceleration of 
1 f/s 2 . Now if a body weighing 1 pound contains m units of mass, 
the equation of motion, f — ma, gives, for the force with which 

gravity acts on the body, \ — m • g, whence m— — ; that is, a body 
weighing 1 pound contains — (about fa) units of mass, or the 



Forces. 273 

unit of mass is the mass of a body weighing g (about 32) pounds 

W 

Consequently, there are — units of mass in a body weighing 11' 

y 
pounds. 

II. In the Metric Gravitational System the unit of force is 1 
gram, the unit of acceleration is 1 cm/s 2 , and the unit of mass, 
to which the unit force gives unit acceleration, weighs 981 grams. 

III. In the English Absolute System the unit of mass is the 
mass of the body (the piece of platinum spoken of earlier) 
which weighs 1 pound, the unit of acceleration is 1 f/s 2 , and the 
unit of force is the force which gives unit acceleration to the unit 
mass. Then, if the force with which gravity acts on a body weigh- 
ing 1 pound (the force of 1 pound) contains / of the units of force 
of this system, the equation of motion, / = ma, gives for the force 
with which gravity acts on this body : 

f=l-g, or f=g; 

that is, the force of 1 pound contains g (about 32) of the units of 
force of this system. The unit of force in the English Absolute 

System is called a poundal; 1 poundal= — pound, about -£% 

pound, or half an ounce. A force of x pounds is thus a force of 
gx poundals — about 32a; poundals; a body weighing W pounds 

W 

or W g poundals contains W absolute units of mass and — gravi- 

tational units of mass. 

IV. In the Metric Absolute System the unit of mass is the 
mass of a body weighing 1 gram, the unit of acceleration is 1 
em/s 2 , and the unit of force is the force which gives unit accelera- 
tion to a unit mass. The absolute metric unit of force is called 

dyne; it is -g^-y gram, or 1 gram = 981 dynes. 

There is no name for any one of the four different units of 
mass ; this is because in practice the mass of a body is always ex- 
pressed in terms of the weight of the body. Thus the equation of 
19 



274 The Calculus. 

motion for a body weighing W pounds, acted upon by a force of f 

pounds, and acquiring an acceleration of a f/s 2 , is /= ■ — a; since 

y 

W 

the body contains — gravitational units of mass. 

Again, a body weighing W poundals, acted upon by a force oi 
f poundals, and acquiring an acceleration of a f/s 2 , has for its 

W W 

equation of motion /' = — a ; since the body weighs ■ — pounds, 

y y 

W 

and so contains — absolute units of mass. 
9 

Consequently, the equation of motion for a body in terms of the 
weight of the body, the force acting, and the acceleration acquired, 

W 

can always be written f= — a, provided that / and W are both 

expressed in terms of the same unit, either the pound or the 
poundal, and that a and g are expressed in terms of the same 
unit, the foot a second each second. 

The same is of course true of the metric units, with centimeter, 
gram and dyne in place of foot, pound and poundal. Indeed, this 

equation, ^~ = — , is merely the assertion that the magnitudes 

of the accelerations given to the same body by two different forces 
are proportional to the magnitudes, f and W, of the forces 
themselves. 

257. Composition and Resolution of Forces. — It is found by 
experiment that when a number of forces act at once on the same 
body, the effect of each is independent of the effect of the others ; 
that is, that the acceleration actually given by all the forces is the 
sum or resultant of the several accelerations that would be given 
by the different forces, each acting by itself. If each of the forces 
acting on a body is represented by a line having the direction of 
the force and having a length proportional to the magnitude of 



Forces. 



275 



the force, all these lines will be in the directions of the accelera- 
tions given by the forces, and their lengths will be proportional to 
the magnitudes of the accelerations. Hence these directed lengths 
will be displacements that represent (to different scales) either 
the forces or the accelerations given by them, and the composition 
and resolution of the forces can be effected by the methods already 
described for displacements, velocities and accelerations. 

That is, the triangle construction, or the corresponding compu- 
tations, can be used to combine two or more of the forces acting on 
a body, or to separate any one of the forces into components acting 
in given directions ; then the effect on the motion of a body of such 
a resultant force is precisely the same as the effect of its com- 
ponents. 

258. It should be observed that this treatment of forces is con- 
cerned merely with directions and magnitudes. A line drawn in 
the direction of a force from the point at which the force is applied 
is called the line of action of the force. Shifting the point of 
application from one point to another along the line of action is 
found to have no effect if the two points are rigidly connected; 
but any other change in the point of application does alter the 
effect of the force. Consequently, in compounding and resolving 
forces a resultant and its components are treated as having the 
same point of application. 

259. Resolved Parts of a Force. — Perpendicular components 
of a force are called resolved parts of the force, as is the case with 



Y 


p 


■//?# 


^0 , \ X 


o 


ycos j? 




Fig. 91. 



Fig. 92. 



£76 The Calculus. 

accelerations. For instance, a force /, which causes an accelera- 
tion a in a given direction OP, may be resolved in two perpen- 
dicular directions, OX and OY , giving, if the angle XOP is called 
<j>, the resolved part / cos </> in the direction OX and the resolved 
part / sin cj> in the direction OY (Fig. 91) . 

Note that the acceleration a in the direction OP amounts to an 
acceleration acos</> in the direction OX and an acceleration 
a sin cf> in the direction OY, and that these are the accelerations 
that would be due to the forces / cos </> and / sin <f> acting indi- 
vidually. 

Again, in order to determine a force / from its resolved parts, 
f x in the direction OX and f y in the direction OY, we have 

'Vfw 2 +fy 2 =f for the magnitude of f, and <j> — tan -1 -'*- = sin -1 Jf- 

h I 

for the direction of /. 

260. Equation of Motion for a Given Direction. — The equation 

W 
of motion, f=ma or /= — a, expresses the relations between 

y 

the magnitudes of f, m. and a for any moving body ; the fact that / 
and a have the same direction is of equal importance, but can be 
no more than implied in the equation. Since, however, a resolved 
part of the acceleration is due to the resolved part in the same 
direction of the force, we can always write the equation of motion 
for the resolved parts in two different directions, thus obtaining 
two algebraic relations which involve both the equality of magni- 
tudes of / and ma and the identity of directions of / and a. The 
Law of Motion is consequently utilized in practice in the following 
form: 

The (algebraic) sum of the resolved parts in any given direc- 
tion of all the forces acting on a body is equal to the mass of the 
body multiplied by the resolved part of the acceleration in that 
direction. 

This is formulated: f ± cos <j> x + f 2 cos</> 2 +/ 3 cos</> 3 + . . . . = 






Forces. 277 

ma, or 2/ cos <j> — ma! ; cf> ± , <f> 2 , <f> 3 , . . . . etc., representing the 
angles between the direction of a and the directions of f 19 / 2 , f 3 , 
.... etc. 

The Equation of Motion may be written for any two directions 
that prove convenient ; there is no reason why the two should be 
perpendicular. It is often desirable to write an equation that shall 
fail to contain one of the forces ; it is evident that this can be done 
by choosing the direction perpendicular to that of the force in 
question. 

261. Examples. 

Use g = 32 unless otherwise indicated. 

1. If 1 pound = 453.59 grams, and 1 meter= 3.281 feet, con- 
vert a pound into dynes, g — 32.19. Ans. 445,000. 

2. If a force of 10 pounds produces in a body an acceleration of 
20 f/s 2 , find the weight and the mass of the body. 

Ans. 16 pounds; -J. 

3. A weight of 10 pounds lies in the scale pan of a spring 
balance, hanging from the top of an elevator ; if the elevator starts 
up with an acceleration of 5 f/s 2 , what would the balance indicate ? 

Ans. 11 pounds 9 ounces. 

4. A weight of 40 pounds is hanging vertically by means of a 
long string ; what force applied horizontally would give the weight 
an acceleration of 4 f/s 2 ? Ans. 5 pounds. 

5. A weight of 10 pounds hanging vertically by a string is 
pushed by a horizontal force so that the string makes an angle of 
30° with the vertical; find the force and the tension of the string. 

Ans. /=5.77 pounds and tension = 11.55 (approx.). 

6. A weight of 100 pounds is suspended from two pegs, placed 
in a horizontal line 5 feet apart, by two cords 3 and 4 feet long, 
respectively ; find the tension in each cord, by construction and also 
by the equations of the forces resolved horizontally and vertically. 

Ans. Shorter cord 80 pounds, the other 60 pounds. 

7. A weight of 2000 pounds is suspended by two ropes making 
angles of 30° and 45° with the vertical, respectively; find t x and 
t 2 , the corresponding tensions in the ropes. 

Ans. ^ = 1464.1,^ = 1035.3 (approx.). 



278 The Calculus. 

8. Two forces of 3 and 5 pounds acting at a point have a re- 
sultant of 7 pounds; find the angle between the forces and also 
between each force and the resultant. 

Ans. 60°, 21° 47' and 38° 13' (approx.). 

9. If a man can lift 180 pounds when standing on the ground, 
how much can be lift in an elevator ascending with an acceleration 
of 8 f /s 2 ? When it is descending with the same acceleration ? 

Ans. 144 and 240 pounds. 

10. An anchor weighing 5000 pounds hangs vertically from the 
end of a boom which makes an angle of 45° with the vertical mast 
where it is hinged ; the outer end is supported by a lift making an 
angle of 60° with the mast; find by resolution of the forces acting 
at the end of the boom, the tension on the lift and the thrust on the 
boom. Ans. T = 3660 pounds, P=4482 pounds. 



CHAPTEE XII. 
Motion op a Heavy Particle. 

262. Definitions. — The equation of motion can be used to deter- 
mine an unknown force when the acceleration is given, or to deter- 
mine the acceleration when the forces are given. In one large 
class of problems it is required that there shall be no motion, i. e., 
that the acceleration, and, therefore, any resolved part of the ac- 
celeration, shall be zero. The study of such problems is called Stat- 
ics. The study of the motion caused by given forces is called Dynam- 
ics. In a problem in dynamics, when the acceleration has been 
determined from the equation of motion, there remains a problem 
in integration, the determination of the velocity and position of the 
moving body at any time. This part of the problem is treated in 
accordance with the principles of Kinematics. We have discussed 
kinematics only so far as concerns the acceleration, velocity and 
displacement of a moving point, and so are prepared to treat only 
those problems in which all the forces may be considered to act at 
a single point, and in which the motion is the same as if all the 
mass of the moving body were concentrated at that point. This 
is not so narrow a restriction as would appear at first sight, for 
as we shall see later, a very large class of problems can be treated 
on this assumption. For instance, we shall prove that the force of 
gravity acts on any body precisely as it would if all the mass of 
the body were concentrated at a definite point within the body, 
called its center of gravity. 

When all the forces acting on a body are considered to act at one 
point, at which the whole mass of the body is concentrated, the 
body is said to be considered as a heavy particle. We shall confine 
ourselves for the present to the Dynamics of a Heavy Particle. 



280 The Calculus. 

263. Rectilinear Motion under Gravity. — A force having a con- 
stant direction produce's an acceleration in the. same constant 
direction ; a force constant in magnitude as well as direction pro- 
duces a constant acceleration in a fixed direction. The simplest 
case of accelerated motion is motion in a straight line with con- 
stant acceleration. As an example of uniformly accelerated recti- 
linear motion consider the motion of a body falling under the 
action of gravity alone. This is either the motion of a body fall- 
ing in a vacuum, or approximately the motion of a body falling 
through the air. 

Suppose the body to be H feet above the ground, moving ver- 
tically upward with a speed of V f/s, and let the weight of the 
body be W pounds. Then the force is W pounds, acting vertically 

W 

downward, and the mass is - — , so that the equation of motion 

W 

gives, if a is the vertical downward acceleration, W= — a, whence 

y 
a=g. 

Then if, at the end of t seconds, the body is s feet above the 

ground and moving v f /s vertically upward, 

as the upward velocity is decreased by the downward acceleration. 
Integrating, we have v=—gt+C. Here C is some constant; 
since, by the given conditions, v = V when £=0, 

V=-g>0 + C, or C = V; 
hence 

v = V-gt (2) 

As s, the distance above the ground, is increased by the upward 
velocity v, v— + -^ . Hence, -^- — Y-gt. Integrating, 
s=Vt-igt 2 + K. 



Motion of a Heavy Particle. 281 

K is an undetermined constant; but by the conditions, s — H 
when t = 0; hence, 

H=V-0-ig-0+K, or K=H. 
Finally, 

s = H+Vt-igt 2 . (3) 

Equations (2) and (3) give the velocity and the position of the 
body after it has been in motion any given number of seconds; 
they apply to the case of a body having an initial velocity down- 
ward, if V is taken negative. 

If we eliminate t from (2) and (3), we derive an expression 
for v in terms of s; the same result can also be obtained by a dif- 
ferent integration of the equation of motion, which is important 
because it is the only feasible method in a large class of problems 
involving variable forces. 

„. dv , ds dv ds dv ,, . i 

Since a = — rr and v= - T - , a = -— = -y • so that we have 
dt dt ds dt ds 

v ■-=- = — g, or vdv— — gds. 

Integrating, ^v 2 = —gs + C, and since v = V when s=H, 

iV 2 =-gH + C; C=i(V 2 + 2gH); 
hence 

v 2 = V 2 + 2g(H-s). (4) 

Thus the increase in the square of the speed is 2g times the dis- 
tance the body has fallen, and, in particular, the speed is the same 
at a given height whether the body is rising or falling. 

264. We have made no use, in the foregoing discussion, of the 
equation of motion for the horizontal direction. The horizontal 
component of gravity is zero, hence the horizontal acceleration is 
zero. As the horizontal velocity is constant, and is zero at the 
start, there is no horizontal motion. In this case, then, the simple 
initial conditions cause the direction of motion to be the same as 
the direction of the acceleration, or of the force. In the example 



282 The Calculus. 

that follows, the force (and, therefore, the acceleration) is ver- 
tical, but the path of motion is a curve. 

265. The Parabolic Trajectory. — Suppose that a body acted 
upon by gravity alone starts from a point H feet above the ground, 
moving with a speed of V f/s at an angle <j> above the horizontal. 
Consider the vertical and horizontal motion separately, with refer- 
ence to axes drawn through the initial position of the body. Then 
if (x,y) are the coordinates of the body at any time, it will have 
moved x feet horizontally and y feet vertically, and will be at 
the height (y+H) feet. The resolved parts of the acceleration 

d?v dv • d?x dx 

and the velocity are -^ and -rf vertically, -rr 2 and -n horizon- 
tally. 




Fig. 93. 

Just as in the simpler case, the equation of motion gives for 
the vertical direction : 

d 2 y 

and for the horizontal direction, 

dx du 

The initial values, however, are V cos <f> f or -^ , V sin <f> for -^ , 

and zero for x and for y. The integration of the equations, there- 
fore, gives 



Motion of a Heavy Particle. 283 

v x =-^ =Vcos<f>, v y =-^- = Vsm<j>-gt; (1) 

x=t - V cos<f>, y=t- V sin</> — \gt 2 . (2) 

The velocity of the body after t seconds is, therefore, 

v= ~Vv x 2 + Vy 2 in magnitude, in a direction making the angle 

tan -1 —*- = cos _1 — — with the horizontal 

v x v 

The curve traced by the body, called its trajectory, is given by 
the pair of parametric equations (2), which furnish the readiest 
means of solving most problems involving the trajectory. Elim- 
inating t, we find as the single equation of the path of motion, 

since t— -~ , 

7cos<£ 

y= ~ gx 9 +a;tan<i>. 
y 27 2 cos 2 (/> v 

The trajectory for a body acted upon by gravity alone is thus a 
parabola. The highest point (x , y ) of the parabola is given by 

dy _ 7 sin<ft — gt _ ft . , _ V sin <f> . 
dx~ 7cos</> ' °~ g ' 

if o 

t i7 • ^ i j z T 72 sin 2 <f> V 2 sin 2 <£ F 2 sin 2 d> 
_ V 2 (l-eos2<f>) 

Transforming to (x , y ) as a new origin, putting x=x + x', 
y=y + y', t = t + t', and dropping primes, we have 

x=t.Yamf, y=-$gt\ or S-jpL* 1 . 
The highest point is, therefore, the vertex; the axis is vertical, 
and the parameter is - — - . 



284 The Calculus. 

In the case of a projectile, fired from a gun, the greatest height 

V 2 

reached, -j— (1 — cos 2<f>), will be a maximum when the elevation 

of the gun is 90°, as this value makes cos 2<f> a minimum. 

The horizontal distance traversed when the projectile regains 
its original level is called the horizontal range; it is evidently 

R = 2x — — — - , and is a maximum when sin 2</> is a maxi- 

if • 

mum, or = 45°. 

266. Rectilinear Motion under any Constant Force. — The 

methods of integration used in determining the motion of a freely 
falling body can be applied directly to any case of rectilinear 
motion under a constant force. For if the constant force is F 
pounds, the equation of motion is : 

W F F 

F= ■ — a, whence a = ^ g, so that the constant ^ g takes the 

place of the constant g throughout the discussion. For instance, 
suppose a body weighing 20 pounds is placed on a board so in- 
clined that if the body is started down the board it will move with 
constant velocity, showing that there is no force acting on it 
(i. e., that the sum of any forces acting is zero), and suppose 
that it is drawn down the plane (starting from rest) by a spring 
balance kept at a constant tension of 2 pounds. Then the equa- 
tion of motion for the direction down the plane is : (<7= 32 f/s 2 ) , 
2 = ff a, whence a = 3.2 f/s 2 . 

Integrating, and determining the constant of integration in 
accordance with the initial conditions, we have for the speed and 
the distance after t seconds : 

v = 3.2t, s=lM\ 

Again, suppose the board on which the weight rests is elevated 
at an angle of 30° to the horizontal. To simplify the problem, 
we will suppose that gravity is the only force acting in a down- 
ward direction along the board, neglecting the effect ', of any 



Motion of a Heavy Particle. 285 

roughness of the board. The resolved part of the force of gravity 
down the board is Wsin 30° = 20x^ = 10 pounds, so that for the 
acceleration down the board we have 

10=ffa, a=16 f/s 2 . 

Hence, t seconds after motion starts, the velocity and distance 
down the board are 

v = 16t and s = St 2 . 

267. Examples. 

1. A weight is dropped from a balloon ascending with uniform 
speed of 20 f/s, and is observed to strike the ground 4 seconds 
later; find the height of the balloon and the velocity with which 
the weight strikes. Ans. 7i = 176 feet, v = 108 f/s. 

2. A steamer approaching a dock with engines reversed, pro- 
ducing uniform retardation, is observed to go 300 feet in 20 
seconds after reversing engines and 100 feet in the next 10 sec- 
onds; find a and v and the distance and time before coming to 
rest. 

Ans. a—— J f/s 2 , v = - 5 g 5 -f/s, £ = 55 seconds, s=504J feet. 

3. A body moves 12 feet while being uniformly retarded from 
24 f/s to 6 f/s ; find the time and acceleration. 

Ans. £ = -f seconds, a= —22.5 f/s 2 . 

4. A projectile fired from the top of a tower at an elevation of 
45° strikes the ground 60 feet from the foot of the tower at the 
end of 4 seconds ; find the height of the tower, also the time before 
striking if the projectile had been fired horizontally. 

Ans. h = 196, £ 2 = 3-| seconds. 

5. A body is thrown at an elevation of 60° with a velocity 150 
f/s ; find the coordinates of its position at the end of 5 seconds, 
and its velocity at that instant. 

Ans. #=249.5 feet, z=375 feet, v= 80.82 f/s, r= -21° 52'. 

6. Show that the time of descent of a body down any chord 
drawn from the highest point of a circle is the same as the time 
of falling down the vertical diameter. 



Ans. t 2 = for all chords. 

9 






286 The Calculus. 

7. A car starts from rest with an acceleration of 4 f/s 2 ; at 
what angle must a man lean forward to keep himself in equi- 
librium? Ans. tan" 1 $ = 7° 8' (nearly). 

8. A train running at 30 m/h against a constant resistance of 
12 pounds per ton of 2000 pounds shuts off steam just as it strikes 
an up-grade of 1 ft. in 200 ft. ; find the time and distance before 
the train comes to rest. Ans. 125 seconds, 5=2750 feet. 

9. Find the distance required in example 8 by direct integration 
of the equation of motion (Art. 266) . 

10. Show that the speed acquired in sliding down a smooth 
plane from rest, under the force of gravity, is the same as for fall- 
ing freely through the height of the plane. 

11. A body weighing 20 pounds is pulled along a smooth hori- 
zontal plane by a horizontal force of 5 pounds. Find its motion. 
How is this motion affected if the pull is inclined at the angle <f> 
to the horizontal ? 

Ans. The body moves with constant acceleration; a=8 f/s 2 , 

a =8 COS <j> f/s 2 . 

12. A body weighing 20 pounds, on a smooth plane inclined at 
the angle tan _1 f to the horizontal, is pulled by a force of 14 
pounds acting up the plane and parallel to the plane. Find its 
motion. How is this motion affected if the pull is inclined at the 
angle 4> to the plane ? 

Ans. The body moves with constant acceleration up the plane : 
a = 3.2 f/s 2 , a' = 3.2(7 cos <j>-6) f/s 2 . 

13. A body weighing 64 pounds, on a smooth plane inclined at 
the angle sin -1 -J to the horizontal, is pulled for 3 seconds by a 
force of 12 pounds acting up the plane and parallel to the plane. 
How far up the plane will it go, and with what speed will it pass 
its initial position in its descent? Ans. 9 feet, 6V2 f/s. 

14. A canal-boat weighing 7-J tons is brought from rest to a 
speed of 3 m/h in one minute by a constant pull of 200 pounds, 
making the angle cos -1 \\ with the direction of motion. Find 
the resistance, assumed constant. 

Ans. 163.4 pounds, making the angle 200° 3' with the direc- 
tion of motion. 

268. Rectilinear Motion under Two or More Forces and under 
Variable Forces. — When a force is applied to a body, it is gener- 



Motion of a Heavy Particle. 287 

ally impossible to prevent other forces from acting, so that the 
motion is caused, not by the applied force alone, but by this force 
combined with one or more resistances. For instance, the motion 
of a body under the action of gravity is affected by the resistance of 
the air through which it falls, by the resistance of a rough sur- 
face down which it slides, or by the resistance of a coiled spring or 
mass of sand upon which it falls. Either an applied force or a 
resistance may vary as it acts. Any force has laws of action, more 
or less accurately determined by experiment, from which its effect 
can be deduced. We shall next consider a few of the most 
familiar forces. 

269. Hooke's law for the force exerted by an elastic rod, cord, 
coiled spring, etc., holds for all practical purposes provided the 
stretching or compression is not great enough to destroy the 
elasticity. It is formulated as follows: 

The natural length being I, and the length when stretched 
l + Al, the force exerted is proportional to Al. 

The actual value of the force depends upon both the material 
and the form of the elastic object. 

For example, suppose a spring 10 inches long, which exerts a 
pull of 2 pounds when stretched to a length of 11 inches, to be 
hung up with its axis vertical, and a body weighing 5 pounds to be 
attached to its lower end. Then when the spring is stretched s 
feet it exerts a pull of 24s pounds. The equation of motion for the 
forces acting on the body in the downward direction (if we neglect 
the weight of the spring and the resistance of the air) is : 

since a = v -=- (see Art. 263). 

Separating the variables, we have: 



288 The Calculus. 

Integrating, and determining the arbitrary constant (5 and 
v are zero together) : 

Again separating the variables, 

dt= -,=£= , or WYidt: 



8Vs-^-s 2 ' ° \Zi\s-s 2 

Integrating, and determining the arbitrary constant, 



^cos-^, (1) 



whence 






1-cos — 5 — J = T 5 2 sin 1 —5 (2) 

The motion is, therefore, an oscillation -£% feet or 5 inches down 
and back, repeated indefinitely. During a complete oscillation, 

cos ^ — goes through a complete cycle of its values, from 1 

through 0, — 1, back to 1, and cos -1 — - — increases from 2&tt 



to 2(& + 1)tt; hence the time of a complete oscillation is 

T= ^ 2* = ^p = .51 second (nearly). (3) 

The neglected resistances will actually shorten the oscillations 
until the body comes to rest 2-| inches below its initial position, 
the spring then being stretched by a steady pull of 5 pounds. 

270. The Force of Gravity. — According to the law of gravi- 
tation (which is universal) any two particles of matter attract 
each other with a force proportional to their masses and in- 
versely proportional to the square of the distance between them. 
It follows from this that a sphere of which the density is the 
same at all points equidistant from its center attracts a particle 



Motion of a Heavy Particle. 289 

at any distance s from its center as if all the mass of the sphere 
were concentrated at its center, with a force inversely propor- 
tional to s 2 if the particle is outside the sphere, directly propor- 
tional to s if the particle is inside the sphere. 

This is a very close approximation to the law of the attraction 
of the earth, or of the force of gravity. If we denote the force 
of gravity on a particle of mass m at a distance s feet from the 
center of the earth by / pounds, the value of / at the surface of 
the earth by mg, and the radius of the earth by a, 

f=-~2 ^ s^>a, f=cms if s<a, f — mg if s=a. 
s 

Hence, taking g = 32.2 f/s 2 , a=3960 mis., b = ga 2 = lAlxl0 1Q , 

c= l- = 1.5ixl0- 6 . 
a 

For a body falling directly toward the center of the earth from 

a great distance, we have, neglecting resistances 

j. bm —b 

f=—T = -ma, or «=-^-. 

For a body falling vertically inside the earth (down a mine 
shaft, for instance) we have, again neglecting resistances, 

j — cms— —ma, or a= — cs. 

These equations are integrated in essentially the same way as 
the equation of motion under Hookers Law. 

271. Examples. 

1, 2. An elastic cord, of natural length I feet, is fastened at a 
point I feet vertically below a hole in a smooth horizontal table. 
A particle P, weighing W pounds, is attached to the free end of the 
cord and placed on the table. Let the particle be s feet from the 
hole at the end of t seconds. 

1. Given W=32 pounds, find the motion if the particle is 
drawn back 3 feet from the hole and let go, the tension of the 
cord being 12 pounds when the particle is started. 
20 



290 The Calculus. 

Ans. 5 = 3 cos 2t; the particle oscillates back and forth, com- 
pleting an oscillation in 3.14 seconds. 

2. Given W=4 pounds, find the motion if the particle is 
started from the hole with an initial (horizontal) speed of 10 
f/s, a steady pull of 2 pounds being needed to hold the particle 
1 foot from the hole. 

Ans. s=f sin 4£; the particle makes a complete oscillation in 
1.57 seconds. 

3. A coiled spring is set up on a firm support with its axis 
vertical; the natural length of the spring is 5 feet; under a steady 
pressure of 10 pounds, the spring is compressed to a length of 4^ 
feet. A weight of 10 pounds is placed upon the spring. Find 
the motion of the weight. 

Ans. If the top of the spring is at under a steady pressure 
of 10 pounds, the weight will be s feet below G, t seconds after it 
has passed G, where s=% sin St; and will oscillate from 6 inches 
below C to 6 inches above and back, making a complete oscilla- 
tion in T=0.79 seconds (nearly). 

4. If the weight in example 3 is dropped upon the spring from 
a point 2 feet above, find the motion. 

Ans. Using s and t as before, s=f sin8#, T=.79 second. 

5. Given that the earth's attraction for a body outside the 
earth is inversely proportional to the square of the distance of 
the body from the center of the earth, show that when a body 
weighing W pounds is s feet from the center, it is attracted with 

Wa 2 

a force of — ^ pounds, the radius of the earth being a feet, and 
s~ 

that if the body is at rest when s=H, its speed is v in gen- 
eral, and v x when it reaches the earth, where 

and the greatest value possible for v ± is less than 7 m/s. 

6. Given that the earth's attraction for a body inside the earth 
is directly proportional to the distance of the body from the 
center of the earth, show that when a body weighing W pounds is 

Ws 

s feet from the center, it is attracted with a force of — - pounds, 

(X 

the radius of the earth being a feet, and that a body starting from 



Motion of a Heavy Particle. . 291 

rest at one end of a hole bored diametrically through the earth 
would reach the other end in n J — seconds = about 42£ minutes. 

272. Resistance of a Rough Surface. — The resistance offered 
to the motion of a body A by a body B with which it is in contact 
is subject to well-established laws. This resistance, the force 
with which B presses against A, is the same as the force with 
which A is pressed against B. The magnitude and direction of 
the resistance, determined from observation of their effect on the 
motion of the resisted body, are found to be subject to limitations 
due to the physical nature of the bodies in contact. 

Thus, if a body is at rest on a firmly supported flat board, the 
resistance of the board is a force equal to the weight of the body, 
directed vertically upward, and as the body may remain at rest 
when the board is tilted, the inclination of the resistance to the 
resisting surface is capable of variation. This variation is lim- 
ited by the roughness of the surface. 

Friction. — The laws governing the resistance of a rough 
surface (commonly called the Laws of Friction) are expressed in 
terms of the resolved parts of the total resistance along and nor- 
mal to the surface ; the normal component R is called the reaction 
of the surface (sometimes the normal reaction) ; the component 
along the surface F is called the frictional resistance or the fric- 
tion (sometimes the tangential reaction). 

If a body is at rest on a surface inclined at an angle 6 to the 

horizon, then, since the resultant of F and R, 

or the total resistance, is a vertical force, 

W 
tan 0= -p- . The greatest angle that will 

JXr 

allow the body to remain at rest is called 
the angle of repose; its tangent is the great- 

F 

est value possible for the ratio -^- . FlG 94 




292 The Calculus. 

The Laws of Friction are : 

I. For any two bodies in stationary contact, the ratio -~ of the 

friction to the reaction is limited ; it can never be greater than a 
certain proper fraction, called the coefficient of statical friction, 
and usually denoted by ft. 

If the angle of repose is a, /x=tana. 

II. The value of ft depends merely upon the roughness of the 
bodies in contact, being independent of the magnitude of the 
total resistance, and of the size and shape of the area over which 
the bodies touch. 

-pi 

III. For any two bodies in moving contact, the ratio -^ is 

constant; its value may be represented by //. ft! is called the 
coefficient of dynamical friction. 

IV. The value of //, like that of ft, depends only upon the 
roughness of the surfaces in contact; it is independent of the 
speed. /A is always less than ft,, but never much less. 

These laws are essentially exact except for very fine points or 
edges and very low or very high speeds. Values of ft and ft! are 
given in engineering hand books for various materials, according 
to their condition of polish and lubrication. 

Two unknown quantities must be found in order to determine 
an unknown force ; the laws of friction determine one of these in 
the case of the resistance of a rough surface if the resisted body 
is in motion along the surface, or at rest and just on the point of 
moving. In the case of a body at rest and not on the point of 
moving, the laws are of no assistance; they are also useless for 
determining the resistance of anything like a rough hinge, for 
which the direction normal to the surfaces in contact is inde- 
terminate. 

273. The following problems will illustrate the use of the laws 
of friction. A body is placed on an inclined plane ; the statical 



Motion of a Heavy Particle. 293 

coefficient of friction for the body and the plane is .45, the dynam- 
ical |; if the angle between the plane and the horizontal is 0, 
find what happens if 0=20°, and if = 30°. 

Since ju=tan a =.45, the angle of repose, a, is between 24° and 
24^°; if 0=20°, the body will, therefore, remain at rest. If 
0=30°, resolve the resistance into the normal and frictional com- 
ponents R and %B and the weight W into the components 

W — W 

W cos 0— — V3 normal to the plane and W sin 6— — downward 

along the plane. If a is the acceleration with which the body 




Fig. 95. 



slides down the plane, the equation of motion gives, for the direc- 
tion down the plane: 

W 2J? W 

and for the direction normal to the plane : 
~ V3-£ = 0. 

From these, a = g U- ^) =.15360. 

The body, therefore, moves down the plane with a constant 
acceleration of about 4.92 f/s 2 , which is independent of the value 
of W. 



294 



The Calculus. 



274. Suppose the body in the preceding example is held from 
moving by a pull directed at an angle <f> with the plane, and sup- 
pose first that the pull P ± is just enough to keep the body from 
sliding down, and next that the pull P 2 is just too little to pull 
the body up the plane. 

In either case, the pull P will have a component along the 
plane, tending to drag the body up, and a component normal to 
the plane which will increase the pressure against the plane if 
the pull is below the plane, and decrease it if the pull is above. 
Consequently, P 1 must be below the plane, and P 2 above it. It 
remains to determine the angle <f> so as to balance to best advan- 
tage the changes in the dragging force and in the friction caused 
by changing <f>. 

In the first case, the resistance points up the plane as much as 
possible, so that the equations for motion along and normal to the 
plane are : 

^-A5R-P 1 cos <f> = 




~ VS-E + P.sin^ 



whence, 



* i 
W 



.1103 



Pig. 96. W cos <f> + .45 sin <f> ' 

In the second case, since the resistance points down the plane 
as much as possible, we have : 




W 

2 

W 
2 



+ A5B-P 2 cos <j> = 
V3-R-P 2 sm<j>=0 



whence, 



P, .8897 

W ~ cos <f> + .45 sin <j> 



Motion of a Heavy Particle. 295 

In either case, the least value of 'P occurs when tan <f> — .45, 
that is, when the angle made with the plane by the pull is equal 
to the angle of repose. The least values are, therefore, 
P x = 0.1006 W, P 2 = 0.8115Tf. These are the limiting values 
desired, for in each case the required pull was the least pull that 
would call into play the maximum frictional component of the 
resistance. 

275. No comprehensive statement exists for the law of atmos- 
pheric resistance. The statement that has generally been con- 
sidered good enough for text-books is that this resistance is pro- 
portional to the projection, on a plane normal to the motion, of 
the surface bounding the moving body (proportional to the " op- 
posed surface") and to the speed of motion for low speeds, the 
square of the speed for medium speeds, and the cube for higher 
speeds. In practice, it has been customary to determine the 
pounds resistance per square foot for a particular type of body 
at varied speeds, tabulating the results. Recent studies of the 
aeroplane have demanded some general law to be used as a basis 
of design, and some progress has been made — enough to show 
that the usual text-book law is by no means general. Sir Hiram 
Maxim has tested aeroplane struts of different cross-sections, and 
found that at a speed of 40 m/h, the resistances of the atmos- 
phere varied from 9.12 pounds to 0.38 pound per square foot of 
opposed surface. We shall need some basis for our problems, and 
so, as it would be inconvenient to give the data in each case, we 
will agree to take the atmospheric resistance proportional to the 
speed, its square or its cube, according as the speed is less than 
20 f/s, between 20 f/s and 500 f/s, or greater than 500 f/s, and 
to call the resistance 7.2 pounds to the square foot of opposed 
surface at a speed of 60 f/s. This is to be understood as merely 
a rule to do problems by ; nothing but experiment with the body 
actually in question will at present give a useful formula. 

The resistance offered by water to the motion of a ship has 



296 The Calculus. 

been more thoroughly studied, but with results that only serve 
to discourage any attempt to invent a simple law, further than 
the obvious one that the resistance increases with the speed of the 
ship until it is equal to the driving force. This is evident be- 
cause the speed of the ship, at first accelerated, soon reaches a 
maximum uniform value. 

276. To illustrate the effect of atmospheric resistance, consider 
the case of a sphere weighing 25 pounds, with a total surface of 
3 square feet, let fall from a height in still air. The opposed 
surface is f square foot, and when the sphere is moving v f/s, the 
atmospheric resistance in pounds per square foot of opposed sur- 
face is .04v if v is less than 20 f/s, .002i; 2 if v is between 20 f/s 
and 500 f/s, .000004?; 3 if v is greater than 500 f/s. Let s be the 
number of feet that the sphere has fallen at the end of t seconds ; 
then s, t and v are zero together. 

While v is less than 20 f/s, the equation of motion is 

25-(.04xf)«=t#a=f!^. (1) 

Separating the variables, we have 1.28 dt= — — — — ; inte- 

grating and determining the arbitrary constant, 

-0.0384* = log(l-0.0012t;) ; (2) 

whence, 

v=~^=833.33(l-e-°- 0S3 * t ). (3) 

Integrating again, and determining the constant, 

s = 833.33£-21,701.5(l-6T - 0384 '). (4) 

277. The law of the resistance changes when ^=20 f/s, 

2 = 0.6326 seconds and 5=6.33 feet. 
In the next stage of the problem, we have 

25-(0.002xfK = M %> (6) 



Motion of a Heavy Particle. 297 

and if s feet is the distance fallen in addition to the fall of 6.33 
feet, the initial values are £ = 0, v = 20, s = 0. 
Treating the equation as before, we have: 

dv 



1.28dt- 25 _ 



0.49574* = log 



awire gg±j- 



(6) 



^" (ft -^ y ' lx i.3667e°-* 9 "« + l' [ } 

s=520.84 log[0.5775e - 24787 ' + 0.4225e-°- 24787 ']. (8) 

If we require the time when the body will have attained a 
given speed, we see that if v is greater than 129.1, formula (6) 
gives an imaginary value to t. But when v — 129.1, the force 
acting on the sphere is from (5), 25 — .0015v 2 = 0, the re- 
sistance at this speed being exactly equal to the force of gravity. 
As a result, the sphere would fall with constant speed from this 
point on; but, according to the formulas, when v = 129.1, t and s 
are infinite. The body falls with diminishing acceleration, -its 
speed approaching the limiting value of 129.1 f/s. At the end 
of 15.22 seconds, v = 129 f/s, 5 = 1679 feet (the sphere has fallen 
from rest to 1685 feet in 15.85 seconds) and the acceleration is less 
than ^o f/s 2 . 

278. Mechanical Connections. — Of somewhat the same nature 

as the topics just treated, is the effect of connecting bodies by 
means of strings or rods that are so light and inelastic that the 
effect of their mass and elasticity may be neglected without 
serious error. Such a rod will transmit a push or a pull without 
altering the magnitude of the transmitted force, merely chang- 
ing its direction and point of application; a string will serve in 
the same way, but of course only to transmit pulls. The direction 
of a taut string is changed by passing it over a pulley, a peg or 
something of the sort, with some loss of force, which, however, 



298 



The Calculus. 



may also be neglected in the case of a light pulley with smooth 
bearings or of a smooth peg. Under these conditions, the two 
tensions in a taut string are directed along the string in both 
directions at any point, are equal and constant throughout the 
length of the string, and so exert the same pull at each end. The 
tension or compression in a stiff light rod is in accordance with 
similar laws. 



279. 



Example. 



Two bodies, A and B, weighing 10 pounds and 25 pounds re- 
spectively, are connected by a light in- 
extensible string. A is placed on an 
inclined plane (// = .2) inclined at an 
angle of 60° to the horizontal; the string 
is passed over a pulley C at the top of the 
plane, so that the part A C is parallel to 
the plane, and the part BC is vertical. 
Find the motion. Let T be the tension of 
the string, F and R the frictional and 
normal components of the resistance of 
the plane,' 1^ = 10 pounds and W 2 = 25 
pounds, the weights of A and B. 

Then, as indicated in Fig. 98, if a is the acceleration of A up 
the plane, or of B vertically downward, the equations of motion 




w 2 =Z5lbs. 



Fig. 98. 



are: 



For A : 



W 
r- J P-Tf 1 8in60° = -^ 1 a l 

9 
22-Tf 1 eos60° = 0, 

F=— . 



And for B 



W 



Thence, 



W. 



T=.2(W 1 cos 60°) + W x sin 60° + — a 



W 2 - 



W. 



S a. 



Motion of a Heavy Particle. 299 



or 

r=9.6603+M a = 25 -M«; 

whence, 

• a = 14.025 f/s 2 . 

The pull of the string is T— 14.043 pounds on each body. 

280. Examples. 

1. Two weights, W 1 and W 2 , are connected by a thin inexten- 
sible cord which passes over a smooth pulley having a horizontal 
axle. Show that the weights move with the acceleration 

2-3. A weight of W t pounds hangs from a thin inextensible 
cord I feet in length which passes over a smooth pulley at the edge 
of a smooth horizontal table and is attached at the other end to a 
weight of W 2 pounds lying on the table. The table top is h feet 
from the floor, and W t starts from rest at the edge of the table. 
Find the motions of the weights. 

2. W ± =4, TF 2 = 21, J=fc=4. 

Ans. W 1 reaches the floor in 1 J seconds ; W 2 reaches the edge 
of the table in 1J seconds and J second later strikes the floor 3-J 
feet from W x . 

3. W 1 = 9 9 TT 2 = 135, 1=12, h = 9. 

Ans. W ± reaches the floor in 3 seconds, W 2 in 4J seconds, 
falling 4r| feet from W ± . 

4. Given the same conditions as in example 2, except that the 
table slopes at an angle of 6 to the horizontal, the edge at which 
the pulley is situated being at the top of the slope and horizontal, 
find the value of if there is no motion, and the acceleration of 
the weights if sin0=^, and if sin0=:f. 

Ans. = sm- 1 ^ T ; 04 = 1.28 f/s 2 , W ± descending; a 2 rr2.56 
f/s 2 , W 2 descending. 

5-6. Given the same conditions as in examples 2-3, except that 
the table is rough and slopes, as in example 4, 9 being tan -1 f ; 
find for what value of (x there would be no motion, and find the 
acceleration of the weights if // = T 5 2 . 

5. F 1 = 32, W 2 = S0. 

Ans. fx must be ^ or greater; a = f^ f/s 2 , W 1 descending. 



300 The Calculus. 

6. W 1 = 4, W 2 = 30. 
Ans. fx must be yV or greater; a=f^ f/s 2 , W 2 descending. 

7. Given the same conditions as in examples 5-6, find for what 
values of W ± there will be no motion if W 2 = 30 and /x— ■J. 

Ans. TF-l between 6 and 30 

8-9. A body weighing W pounds rests on a rough horizontal 
plane and is pulled by a force of P pounds directed at an angle 
cj> above the plane. 

8. W=15, fi=%. Find P if the body is just on the point of 
moving, and (f> = 0; again, <£ = tan _1 T 5 2-, <£ = tan -1 f. For what 
value of <f> is P least, and what is the least value of P ? 

Ans. P 1 = 6, P 2 = H, P s = 5{h cA-tan- 1 !, 2%= 5.57. 

9. W = 12, //=£, P=15. Find the acceleration, a f/s 2 , with 
which the body moves according as <f> = 0, tan -1 f ; the value of <f> 
for which a is greatest, and the greatest value of a. 

Ans. a 1 = 29J, a 2 = 29^, ^tan" 1 J, a 3 = 31.49. 

10-11. A weight of W t pounds rests on a rough horizontal 
table-top, h feet above the floor ; a light inextensible string I feet 
long is attached to the weight at one end, passes over a smooth 
pulley at the edge of the table, and is attached at the other end to 
a weight of W 2 pounds, which hangs without other support, 
Find the motions of the weights if at the start the string is taut 
and W 2 is at the edge of the table. 

10. 1 = 6, h=6, T7i = 10, F 2 = 5, // = f. 
Ans. W 2 reaches the floor in 2.37 seconds; 0.62 second later 

W ! strikes the floor 3.1 feet from Wa- 
ll. 1=10, h=6, W ± = 15, W 2 = 9, // = -!. 
Ans. W 2 reaches the floor in 1-J seconds; f second later W 1 

stops 1 foot from the edge of the table. 

12-13. The ridge of a peak roof is h feet higher than the eaves 
and the breadth is 2b feet'between the eaves. Two weights, oi 
W x pounds and W 2 pounds, one on each slope of the roof, are 
connected by a taut rope I feet long, which passes over a smooth 
pulley at the ridge in a plane perpendicular to the ridge. 

12. Wi = 120, W 2 = 300 and W ± is just about to slide up. Li 
h = 10, 1 = 15, what is fi ? 

Ans. /*=f. 



Motion of a Heavy Particle. 301 

13. Wi = 53, W 2 = 331, fi=i, h = 9, b = 12. Find the motion 
3f the weights, given that 1=15, that W 2 starts at the ridge, that 
the rope slips from both weights when W ± reaches the ridge, and 
that the eaves are 3 Of feet above the ground. 

Ans. W t reaches the ridge in 2 seconds, and 1$ seconds later 
strikes the ground 10-J feet from the building; W 2 reaches the 
aves in 2 seconds, and 1-J seconds later strikes the "ground 13-J 
feet from the building. 

14. A sphere d feet in diameter, weighing w pounds to the 
ubic foot, falls from rest; show that until its speed becomes 20 

f/s, its acceleration is — (c — v), where c= -JH- , fc=0.04. 
c v ' 3k 

Show that a rain-drop 0.12 inch in diameter cannot acquire a 
speed of 10 T 5 2- f/s in a vertical fall, given w=62.5 

Ans. t= — W — ,c=W. 

g to C — V lz 

15. If the initial velocity of the sphere in example 14 is ver- 
tically downward, and between 20 f/s and 500 f/s, show that until 

its speed becomes 500 f/s, its acceleration is -^ (a 2 — v 2 ), where 
L 2^, fc = 0.002. 

Show that bird-shot 0.028 inch in diameter fired vertically 
upward in still air will fall to the ground with a speed of less than 
23 J f/s, given w= 700. 

281. The Simple Pendulum. — The simple pendulum consists of 
a heavy particle suspended from a fixed point by a weightless in- 
extensible string or rod, and moving in a vertical circle. 

Let O be the fixed point, let the particle start from rest at A, 
and reach the position P after t seconds. Call the lowest point 
of the circle B, and let angle BO A = o , angle BOP=0, arc BP = s. 
Let the weight of the particle be W pounds, and call the length 
of the pendulum OP = l feet. 



302 



The Calculus. 



The motion of the particle is caused by a tension N directed 
along PO, and the weight W directed 
vertically downward. For the motion 

W 

in the path at P we have — at — W sin 0, 




since N has no tangential component. 
at --~dt> 



d 2 s la 



hence 



at 



Fig. 99. 



d 2 s 
dt 2 

dO 



-I 



d 2 
dt 2 



■ g sin 0. 



If we call — r- = to, we have 
dt 



and 



d 2 
dt 2 

d 2 
dt 2 



dco 
dt 



dw dO 
~dO ~dt 



duo 
~dO' 



du> 
~dO 



^r = — r- sin 0. 



Integrating, we have, since <o = when 0=6 O , 

Oi 2 —-3-{ cos 6 — cos 6 ) . 
Hence 

M l~W~t a T\ 

°>=-jT =— Y —r-(cos0— COS0 O ), 

-de 



Jf*. 



The function 



Vcos0— cos 6 

dt) 

is a new function somewhat like 



Vcos0— cos 9 



the j V 1 — e 2 cos 2 <j>d^> that we found in Art. 205 for the length of 
an elliptical arc. 

282. If O is small, we can get an approximation to the time 
required to swing through a given angle by putting 



Motion of a Heavy Particle. 303 

cos0 o = l-^, cos 6=1-^-, or cos 0-cos o = J(0 o 2 -<9 2 ). 

z z 

We then have, since £ = when 0=0 O , 

/J , f -de e 

vf' Hv^ =cos *;- 

An inverse cosine has any number of values, but since t starts 
from and increases gradually throughout the motion, cos -1 -^— = 

a/ — • tf must do the same, starting from cos -1 -^° .= when the 



particle P is at A, becoming cos -1 -w- = -=■ when P reaches B, 

"o ^ 
n 

cos -1 -^— ° = 7r when P reaches A'. The motion from A to A' is 
called a vibration; at the end of the vibration, 

o» = ^=-^(cos(-0 o )-cos0 o )=O, 

and the particle starts from A' under the same conditions as it 
left A. The pendulum therefore vibrates back and forth between 
A and A 1 , and if T is the time in seconds of a vibration, 

J^-T — tz, T — tt a/ — seconds (approximately). 

283. If O is not small enough for this approximation to be 
sufficiently accurate, we can obtain any required degree of ap- 
proximation as follows: lTi\J~fT= . write 

y I Je VCOS0 — cos O ' 

8 9 

cos 0=1 — 2 sin 2 -— , cos o = l — 2 sin 2 -^ ; then, noting that the 
Z A 

values of the integrand are the same for equal and opposite values 

of 6, we have : 



304 The Calculus. 

dO 



v l Jo vX 8in '"2 
T-W-r 



sin 2 4. 



K ■ 2 6 

sin*-sr — sin-'-g- 
To simplify the limits, put sin-^- = sin-# sin<£; then 



WE^=^?i 



d<£ 



^» cos i_- * ^i-sin'Asin** 



Abbreviate sin -^- by k; then 



n 



Jo Vl-A; 2 , 



V g Jo Vl-& 2 sin 2 <£ 
Since 

(1-& 2 sin 2 </>)-* = l + p 2 sin 2 <£ + il? A; 4 sin 4 <£ 

Zi ' 4 

and 

r= ,;i[ 1+(J) . i . + ( H j l . +( |i f |j l ._ 

284. Taking # = 32.16 f/s 2 , tt = 3.1416, the first approximation 
T = xJ gives as the length of the second's pendulum, 

(T = l) 1= 4- = 39 - 10 inches. 



Motion of a Heavy Particle. 305 

For the second approximation, this result must be divided by 

If O = 5° (it is never more for an accurate clock), 
^-= i sin 2 (2° 30') =0.0004757, 

so that the more accurate value of I is 

39.10 x (1-0.00095) =39.06 inches. 
The effect of the third term of the series is inappreciable. If a 
pendulum -J^ feet long vibrates in 1 second, a pendulum -^- 

feet long vibrates in 8 seconds and in one day makes 86,400-=-$ = 
86,400(l + 0.000476)" 1 vibrations; hence a clock with a pendu- 
lum constructed according to the formula ir 2 l = g will lose about 
41.1 seconds a day if it swings over a complete arc of 10°. 

285. Variation of Gravity. — In this discussion of the simple 
pendulum, W, the weight of the moving particle, is the earth's 
attraction for the particle at mean sea-level in London (latitude 
51°) and the acceleration given by this attraction at the same 
place is 32.19 f/s 2 . These are the standard conditions mentioned 
in Art. 256 as the basis for comparing force, acceleration and mass. 
The earth's attraction for a body at mean sea-level varies with the 
latitude, and the acceleration it gives to the body varies in the 
same ratio. The values at sea-level for a few latitudes are : 

L= 0° 40° 51° 90° 

g = 32.09 32.16 32.19 (=g) 32.26 

logg = 1.50637 1.50729 1.50772 1.50860 

The acceleration due to gravity also varies with the distance 
above or below sea-level, being f j- J g at a place h feet above 

sea-level, and (^-— ) 9o at a place d feet below, a being the earth's 

radius in feet. 
21 



306 The Calculus. 



The earth's attraction for a body weighing W pounds is 

W 

mg' = g' , where g is the acceleration due to gravity at the 

place where the body is situated, and # = 32.19. Then, in the 
problem of the simple pendulum, the equation of motion becomes 

w w 

— - a t = q' sin 6; or a t = q f sin 6, 

g g 

instead of at — g sin 0, so that g' takes the place of g throughout. 

286. Examples. 

1. What must be the value of o for a pendulum of length 
1= -j-to lose 1 second in a day? Ans. O = 46' 47". 

2. A pendulum that beats seconds when swinging through a 
very small arc is made to vibrate through an arc of 120° 
(0 O :=6O ) ; how many seconds will it lose in a day? 

Ans. About 5890 seconds = l h 38 m 10 s . 

3. A pendulum constructed to beat seconds in a locality where 
# = 32.19, is found to lose 72 seconds a day. What is the value 
of g where the pendulum is, and what change must be made in its 
length to adjust it? 

Ans. #' = 32.137; length must be decreased 0.165 per cent. 

4. If the pendulum of example 3 is on a mountain in latitude 
40°, how high is the mountain? (Take the earth's radius 4000 
miles.) 

Ans, About 1.4 miles. 

5. A pendulum beats seconds at sea-level; how many seconds 
a day will it lose at the bottom of a mine 1056 feet deep ? 

Ans. 2.16 seconds. 

6. A spring-balance is graduated in London at sea-level; what 
will it read when sustaining a 10-pound weight at sea-level at 
the equator? Ans. 9.969 pounds. 

7. Show that the weight given by a lever-balance is inde- 
pendent of the variation of gravity. 

287. Motion in a Vertical Circle. — If the pendulum of Art. 
281 is started by giving the particle P an initial speed of v f/s 



Motion of a Heavy Particle. 307 



v 
in the circle, or an initial angular rate, w = -y , the first inte- 
gration of the equation of motion gives 

(o 2 =:(O 2 -\ — f- (cos 6 — cos O ). 

The force N is determined from the equation of motion in the 
direction PO : 



whence 



J\ — W COS 6— a n = j- = 

g gig 



-^- =C os0+-— =J^L- + 3cos0-2eos0 o . 
W a a 



lio 2 

to 2 becomes zero when 6 — 6 19 if cos $ x = cos o — 



2g 
la 2 

N becomes zero when 9=0 2 , if cos 2 — t cos #o o^~ ~i cos ^i- 

B x and 6 2 are in the same quadrant ; if they are acute, 6 2 is greater 
than X ; if obtuse, d 1 is greater than 2 . 

1 2 
Hence if -^- <cos O , the upward motion of P will cease before 

y can become zero, and the pendulum will vibrate. 

7 2 

If — r °- >cos o , N becomes zero before w = 0. If PO is a rod, 

it will keep moving until w:=0, in the meantime being under 
compression, and then will vibrate back. If PO is a string, how- 
ever, P will start from the point (I, 2 ) with an initial velocity 
in a direction making the angle -n — 6 2 with the horizontal, and 
will move under the action of gravity alone (in a parabola) until 
stopped by the string.. 

7 2 
If -^- > 1 + cos , it is impossible for w to be zero, and if PO 

- 2g 

is a rod, the pendulum will make complete revolutions. 

\w 2 

If -j^- >§ + cos0 o , it is impossible for N to be zero, and the 



308 



The Calculus. 



pendulum will make complete revolutions whether PO is a string 
or a rod. 

288. In the problems so far discussed, we have been required to 
find the nature of the motion from data concerning the forces; 
in the next article we shall consider an example of the. inverse 
problem of finding the relations of the forces when the motion is 
completely given. 

289. Motion in a Horizontal Circle. — If a particle acted upon 
by gravity is constrained by any resistance of R pounds to move 




Fig. 100. 



in a horizontal circle of radius a feet, with a uniform speed of 
v f/s, the resultant of R and the weight of the particle, W pounds, 

must be a horizontal force — pounds, directed toward the 

g a r 

center of the circle, for the total acceleration of the particle is 

— in this direction. Suppose the line of action of R to be PO, 

and the vertical through the center C of the circle to be OC; let 
the displacement PO represent the force R; then the displace- 

Wv 2 

ments OC and PC must represent W and, . If is the angle 



Motion of a Heavy Particle. 309 



v 
made by R with the vertical, tan = , and R — W sec in 

ga 



magnitude and is always in the same vertical plane as PC. 



In applications of this principle, the angle = tan _1 — is 

usually the only important value. For instance, along a railroad 

curve a feet in radius, over which trains pass at v f/s, the road- 

v 2 

bed should have a lateral slope of tan -1 to the horizontal, so 

ga 

that the force R may be exerted wholly against the tires of the 

wheels rather than partly against the flanges. 

Again, if the rods of a centrifugal governor are I feet in length, 

and if o> is the greatest angular speed that the balls should have, 

the governor is adjusted to shut off steam when the rods make 

the angle 6 with the vertical, where 

, „ v 2 aoj 2 I sin &o 2 A n 

tan 6 = = = , or cos ■ 



ga g g ' W 

In the same way, it is found that if we wish to cause the 
particle in Art. 281 to move from A in a horizontal circle, we 
must give it the initial velocity v = lu = y/ gl sec# f/s in the 
horizontal direction perpendicular to AO. When the simple pen- 
dulum moves in this way it is called the conical pendulum. 

290. Examples. 

1. A lead is swung in a vertical circle of 30 inches radius with 
just enough force to make complete revolutions. What is its 
speed at the top of the circle, at the bottom, and when it is let 
go so as to start off at an angle of 45° to the horizontal ? ((7 = 32.) 

Ans. 8.944 f/s, 20 f/s, 18.792 f/s or 11.267 f/s. 

2. A curve on a railway of 4' 8" gauge has a radius of 847 feet. 
How much must the outer rail be raised above the inner for trains 
going at the rates of 15 m/h, 30 m/h, 45 m/h and 60 m/h ? 

Ans. 1 inch, 4 inches, 9 inches, 16 inches, 

3. The inner edge of a semi-circular curve on a running track 
has a radius of 105 feet, the outer edge a radius of 121 feet; the 



310 The Calculus. 

track is flat and banked so that the outer edge is 2 feet higher 
than the inner. For what range of speeds is the track adapted ? 

Ans. From 20J f/s to 22 f/s. 

4. What angle will be made with the vertical by a simple 
pendulum hung in a car which is going 60 m/h around a curve 
of 1000 feet radius? Ans. tan" 1 0.242 = 13° 36'. 

5. If the gauge of the track in the preceding example is 4' 8" 
and the outer rail is 7" higher than the inner, what must be the 
coefficient of friction between a box and the floor of the car if 
the box is not to slide? What if the two rails are on the same 
level? Ans. .114, .242. 

6. A particle is placed at the top of a fixed smooth sphere and 
slightly displaced. Show that it will leave the sphere when it 
has descended over an arc = cos -1 f, and has acquired a speed of 
iVGga f/s, a being the radius of the sphere. 






CHAPTER XIII. 
Momentum, Work and Energy. 

291. General results of great importance can be obtained from 
the equation of motion, / = raa. In the discussion to follow, / 
may be the resultant of all the forces acting, so that a is the total 
acceleration, or / may be the resultant of the resolved parts of 
all the forces in some given direction, and a the resolved part of 
the acceleration in the same direction. 

292. Mean Speed under Constant Force. — From the relations 

, dv ds 

f=m*, a= w , v=- 3r , 

it is readily seen that a constant force / will move the mass m a 
distance of s feet in t seconds, changing its speed from v to v lt 
where v x = v + at, and s— (v -\-%at)t. 

The mean speed V, at which the same distance would be cov- 
ered in the same time, is 



s . ., , v + v. 



and is equal to the average of the initial and final speeds. 

This relation also holds for curvilinear motion under a force 
having a constant tangential component. 

293. Momentum. — The product, mv, of the mass of a body by 
its velocity, is called its momentum in the direction of the velocity. 
Momentum is thus a directed quantity, bearing the same relation 
to velocity that force bears to acceleration. Momenta can there- 
fore be compounded and resolved by the methods given in Chap- 
ter X, and the relations between momentum and velocity treated 
in the same way as those connecting force and acceleration. 



312 The Calculus. 

From the relations f = ma=m — ^- = ^f ■ > it appears that 

the force acting to produce motion in a given direction is the 
time-rate of the momentum, in that direction; that is, force is 
related to momentum as acceleration is related to velocity. 

We may observe that / = ^ _ ' is the equation of motion for 

a body with variable mass. 

294. Impulse. — Calling the momentum mv = M, 

dM=d(mv) =fdt, 

so that if the force / changes the momentum from M to M x in t 
seconds, the total change in momentum is 

M ± -M = \ Ml dM= P fdt. (1) 

JM JO 

This value is called the impulse given by the force / in the time t 
In case the mass m is constant, the total change in momentum 
(or the impulse) is 

mfa-vj^fdt; (2) 

and if the force / is also constant, — F, we have, as a check, 

m(v x — v )=Ft, or v x — v = at i (3) 

where a— — is constant. 
m 

The mean value of the force 




or to the constant force which would produce the same change of 
momentum (the same impulse) as / in the same time. 

Equation (1) or (2) can be used to compare the effects of one 
force on two different bodies, even when the force itself and the 
time during which it acts are both unknown. For instance, sup- 



Momentum, Work and Energy. 313 

pose a 1-ounce bullet is fired from a 15-pound rifle, leaving the 
muzzle with a speed of 2000 f/s, with what initial speed does the 
rifle recoil? Here the gaseous pressures against the bullet and 
against the end of the chamber are the same, though the intensity 
of the pressure is unknown ; so that if we neglect the resistances 
to the motion of the bullet through the bore, we have for the 
bullet and for the rifle 

^(2000-0)= JWl|f(„_0) = j;R 

and / and t are the same in the two equations. 

Consequently, the rifle recoils with a speed of v — -zr= X 



15 ' N 16X32 

= 8J f/s. If the recoil is stopped in T V second, the mean resist- 
ance is 

^=if X-¥X-^ = 39 T V pounds. 

Suppose the gun recoils 2 inches; then, if the retarding force 
were constant, the average speed would be -^ f/s, and the recoil 
would take AXxo = 2V second. The mean resistance is therefore 

HX¥X¥=97.7 pounds. 

295. Impact. — As a further example of the use of momentum, 
consider what happens when two bodies collide. The pressures 
which the two bodies exert on each other change the velocities 
of the bodies as any forces do, but the time during which they 
act is often so short that it is impossible to make any observations 
for determining the rates of change of the velocities (the accelera- 
tions) or the rates of change of the momenta (the forces) . Con- 
sequently the equation of motion, f — nia, is of no use. 

The total changes in velocity and the total changes in momen- 
tum (the impulses) can be observed, however, and the laws that 
govern them can be used to determine the effect of a collision or' 
impact. 



314 The Calculus. 

The pressures exerted on each other by two bodies in collision 
are equal in magnitude and opposite in direction and act during 
the same interval of time; consequently, the impulses or changes 
in momentum are equal and opposite, that is, the sum of the 
momenta of the two bodies is unchanged by the collision. In- case 
one of the bodies is so large that its change of motion is too small 
to be observed, this relation, though still true, is useless — as, for 
instance, when one of the bodies is " fixed " ; i. e., cannot move 
without moving the earth. 

It is found by experiment that the same laws of friction hold 
for impulses as for forces ; we can treat here, however, only cases 
in which friction is negligible. We shall consider the impact of a 
particle impinging on a smooth fixed surface, and the impact of 
two smooth particles moving in the same straight line. In these 
cases, a homogeneous sphere can be treated as a particle. 

Experiment shows that the effect on two bodies of the impulse 
of their collision is to change their relative velocity by reversing 
its component in the direction of the impulse and multiplying the 
magnitude of this component by a number e, called the coefficient 
of restitution, the value of which depends on the elasticity of the 
bodies. The perpendicular component of the relative velocity is 
unchanged. 

296. For instance, suppose a body weighing 12 pounds moves 
10 f/s in a direction making an angle of 30° with a smooth fixed 
surface, and suppose the coefficient of restitution for the body and 
the surface to be f ; find the motion after the body strikes the 
surface. 

In this case, since the surface is fixed, the relative velocity of 
the body and the surface is the velocity of the body. Since the 
surface is smooth, the impulse is normal to it; the component of 
the velocity normal to the surface is 5 f/s toward the surface, 
and is changed by the impulse to fxS^- 1 / f/s away from the 
surface; the component of the velocity along the surface is un- 



Momentum, Work and Exergy. 315 

changed by the impulse. Hence the body leaves the surface in a 



_i_o 

3 



-i 2V3 _ 



direction making the angle tan -1 -~= = tan -1 — ^— =21° 3 



with the surface, at V(5V3) 2 + (^) 2 = 9.28 f/s. 

Suppose that a ball weighing 12 pounds and moving 10 f/s to 
the right meets a ball weighing 3 pounds and moving in the same 
line 15 f/s to the left, and suppose the coefficient of restitution 
for the balls to be f. 

To avoid confusion about direction and sign in a problem of this 
sort, it is well to fix upon one direction as positive, and to express 




Fig. 101. 

all velocities accordingly. The relative velocity must be found 
by subtracting the velocities of the two balls in the same order 
before and after impact. Taking the positive direction to the 
right, we have, if after the impact the 12-pound ball moves v x f/s 
and the 3-pound ball v 2 f/s (each to the right) : 

Hv 1 -hA^=«xio+A(x-i5)=|f > 

since the sum of the momenta is unchanged, and 

»!-»,= -t[10-(-16)] = -H* 

since the relative velocity is reversed and multiplied by e. These 
equations, 

4v, 



:V 1 + V 2 = 25 

© 1 -v 2 =-20| 



316 The Calculus. 

give v ± =l f/s, v 2 — ^l f/s, both balls moving to the right after 
the collision. 

If these balls had been inelastic (e = 0), we should have had 
v 1 — v 2 — 0, v\ = v 2 — h f /s to the right ; if e had been 1, we should 
have had v t = 0,v 2 = 25 f /s to the right. 

297. Examples. 

1. A particle impinges upon a smooth surface, moving along a 
line that makes the angle a with the surface, and rebounds along 
a line making the angle /? with the surface. Prove that tan/? 
= e tan a. 

2. Show that if a billiard ball is knocked around the table, its 
path across a corner is parallel to its path across the opposite 
corner. 

3. A ball A, weighing 5 pounds, moving 7 f/s, is struck by a 
ball B, weighing 6 pounds, moving in the same direction; after 
impact the speed of A is doubled. e—%. Find the speed of B 
before and after impact. Ans. 14 f/s and 8 J f/s. 

4. Two balls, moving 25 f/s and 16 f/s in opposite directions, 
impinge directly upon each other. e = f. Find the distance 
between them 4J seconds after the collision. Ans. 123 feet. 

298. Work. — If the point of application of a constant force F 
moves the distance s in the direction of the force, the product Fs 
is called the work done oy the force. If the point of application 
of a variable force / moves through the distance s in the direction 

of the force, the work done by the force is fds. In either case, 

if the motion is along a path of length s at an angle <f> to the 

direction of the force, the work done is / cos <f> ds. 

Jo 
It is evident from this definition that the work done by a re- 
sultant force is the sum of the work done by its components; if 
each of the forces acting on a particle is resolved along and per- 
pendicular to the motion of the particle, each of the normal com- 
ponents will clearly do no work (cos <f> = 0), so that all the work 
is done by the components in the direction of motion. This is 



Momentum, Work and Energy. 317 

also evident directly from the definition, since if / represents any 
force acting on the body, / cos <f> is its component in the direction 
of motion. 

The work done by a force is positive or negative according as 
the force has a positive or negative component in the direction in 
which its point of application moves. When a force does negative 
work on a body, the body is said to do positive work against the 
force. 

The unit of work, in the English Gravitational System, is the 
foot-pound, the work done by a force of 1 ponnd as its point of 
application moves 1 foot in the direction of the force. 

299. As an illustration of the definition of work, consider the 
work done by gravity when a body moves along the sides of a 
right triangle ABC, of which BC is ver- 
tical and AC is horizontal (Fig. 102). 
If the body weighs W pounds, f=W 
pounds. If the motion is from B to C, 

the work done is directly Wa foot-pounds, /? "^ tt 

)a 
/ cos <fids, f=W pounds, <f> = 0. Fig. 102. 

o 
If the motion is from C to B, <f>, the angle between the downward 
direction of gravity and the upward direction of motion, is 180°, 

so the work is — Wds=— Wa foot-pounds. If the motion is 
Jo 

from C to A or from A to C, ^ = 90°, and the work done by 
gravity is zero. In the motion from B to A, <f> = cos -1 — , and 

(c a 
W — ds — Wa foot-pounds ; in the motion from 
o c 

A to B, = 180° -cos- 1 — , and the work is f * - — ds= -Wa 
^ c Jo c 

foot-pounds. In any circuit of the triangle, gravity does as much 

negative work as positive, or a total of zero. 




318 The Calculus. 

300. Work Done by a Constant Force. — If the point of appli- 
cation of a constant force / moves in a curve, let the axis of x be 
drawn in the direction of f; then, in the expression for work, 

/ cos <f>ds, <f> is the angle made by the direction of motion with 



: 



the axis of x, and cos <b—-^- . Therefore, if the abscissas of the 
^ ds 

initial and final points of the path are x and x, the work is 

/cos</><is = fdx—(x — x )f. 
Jo J«o 

The work is therefore independent of the path traversed, 
depending only upon the magnitude of / and the component in 
the direction of / of the displacement given to its point of 
application. 

301. Work Done by a Central Force. — A central force is always 
directed toward a fixed point and varies in 
magnitude according to the distance of its 
point of application from the fixed point. 
Let be the fixed point, and A the point of 
application of the central force / (Fig. 103) . 
Then if A moves over any path, the angle <f> 
between the direction of the force and the 
Fig. 103. direction of motion is the angle between the 

dv 
radius vector and the path ; hence cos </> = -7- . But by definition, 

the force is a function of r, say F(r). Then the work done as A 
moves along the path is 




S f cos <f>ds=\ F( 
Jro 



r) dr, 



and depends merely upon the limits of integration, and the initial 
and final distances of A from O, being otherwise independent, of 
the path of motion. 






Momentum, Work and Energy. 319 



302. Work Done by Gravity. — The force of gravity is very 
nearly constant for differences of level less than a thousand feet, 
and is for all levels practically a central force directed toward the 
center of the earth. Thus, if a body weighing W pounds is low- 
ered h feet, the work done by gravity is hW foot-pounds, no matter 
what path the body traverses to reach the lower level. Moreover, 
although at present the application of this law is apparently 
limited by the assumption that the body can be regarded as a 
particle, we shall see later (Art. 323) that in this connection any 
body can be treated as a heavy particle. 

303. Examples. 

1. A body, weight 6 pounds, starting from rest, is drawn by a 
cord up a rough inclined plane in 3 seconds, the cord being 
parallel to the plane and the motion uniformly accelerated. In- 
clination of plane 30°, length 10 feet, fi=i- Find the tension 
of the cord and the work done by it. 

Ans. T — 4.456 pounds. Work = 44.56 foot-pounds. 

2. An elastic cord is stretched by a gradually increasing pull 
until its natural length of 1 foot is increased to 18 inches, the 
pull then being 24 pounds. Find the work done by the pull, and 
the work that would be done by gravity in the first descent of a 
weight of 24 pounds hung on the end of the cord, the other end 
being supported. (See Art. 269.) 

Ans. 6 foot-pounds and 24 foot-pounds. 

304. Work and Energy. — If a particle of mass m moves in t 
seconds over a path of s feet, the speed changing from v f/s to 
v f/s under the action of a force of F pounds whose component in 
the direction of motion is F cos (f>=f pounds, then since 

dv 
f=ma=m w , 

we have 

fds — m -rr ds = m —j- dv — m vdv. 
at at 



320 The Calculus. 

and the 

Work = fds=\ mvdv = im(v 2 — v 2 ). 
Jo ]v 

Here / is positive if it accelerates the motion (<j> acute), nega- 
tive if it retards the motion (<f> obtuse) . 

The product, \mv 2 , of the mass of a body by one-half the square 
of its speed, is called the kinetic energy of the body. The change 
in the kinetic energy of a body is the same in magnitude and sign 
as the work done in changing the motion of the body. The 
kinetic energy of a moving body is thus the amount of negative 
work that must be done on the body to bring it to rest, or is the 
amount of work that the body, by virtue of its motion, can do 
against a resistance. The value mv 2 , twice the kinetic energy, is 
often called vis viva, or active force. 

305. For instance, if a constant pull of 12 pounds, exerted on a 
body weighing 9 pounds, lifts the body 3 feet, the pull does 36 
foot-pounds of work on the body, gravity does —27 foot-pounds, 
and both together 9 foot-pounds. Or, the total work is done by 
the upward resultant of 3 pounds as its point of application moves 
3 feet upward. 

The forces therefore make a change of 9 foot-pounds in the 
kinetic energy of the body, so that 

9=i m (v*-v 2 ) =&(v 2 -v 2 ) ; v 2 -v 2 = te. 

Starting from rest, the body would acquire an upward speed 
of 8 f /s, and if the pull were then released, it would rise by virtue 
of its 9 foot-pounds of kinetic energy, going up 1 foot against the 
action of gravity. In falling 4 feet back to its original level, the 
body would regain through the action of gravity the 36 foot-pounds 
of energy given to it by the upward pull. 

Starting with an upward or downward speed of 6 f/s, the body 
would acquire an upward speed of 10 f/s, and if the pull were 
then released would rise 1-ft- feet higher through its \ 2 / foot- 



Momentum, Work and Energy. 321 

pounds of kinetic energy, and in falling to its original level 
would regain 41^ foot-pounds of kinetic energy, its initial 5-j^ 
foot-pounds plus the energy given to it by the upward pull. 

Again, suppose a body weighing 12 pounds is projected along 
a rough horizontal plane (// = ■£) with an initial speed of 8 f/s. 
The body has if X 64 foot-pounds of kinetic energy, and is acted 
upon by a retarding force of -^- = 4 pounds, which will do —12 
foot-pounds of work, reducing the energy to zero, while the body 
moves 1 ^-=3 feet. The body therefore comes to rest after it has 
moved 3 feet and has done 12 foot-pounds of work against a 
resistance of 4 pounds. 

306. Potential Energy. — There is an important physical dif- 
ference in the nature of the forces in these two examples. In the 
first example no energy is lost, for the body has at any time, 
either in kinetic energy or through its position, the power of doing 
as much work as has been expended on it, and returning to its 
original level with its original kinetic energy left. The work it 
can do through its position is called potential energy; forces like 
those of this problem, which do no work that is not available in 
the form of either kinetic or potential energy are called conserva- 
tive forces. The resistance of the rough plane in the second 
example is not a conservative force; the kinetic energy that it 
takes away is lost as far as mechanical effects are concerned. 

If a body weighing W pounds rises h feet, its kinetic energy 
is decreased Wh foot-pounds by the action of gravity. At the 
same time, its potential energy is increased Wh foot-pounds, for 
in falling h feet the body would acquire a speed of \/2gh f/s and 

therefore-^— 2gh=Wh foot-pounds of kinetic energy. Gravity 

thus causes no change in the total energy of a moving body, so 
that the change effected in the total energy by the combined action 
of gravity and any other forces is equal to the work done by the 
other forces. 
22 



322 The Calculus. 

307. Examples. 

1. A 400-pound shot is fired with a muzzle velocity of 1800 f/s 
from a gun weighing 50 tons; find velocity of recoil and the ratio 
of the kinetic energy of the gun to that of the shot. 

Ans. 6f f/s, 2M. 

2. Prove that when a shot is fired from a gun, the kinetic 
energies of the shot and the gun are inversely proportional to 
their weights. 

3. Determine the mean effective pressure in the bore of a 
6-pound gun if the projectile travels 7 feet before leaving the 
muzzle at 2100 f/s. Ans. 59,062-J pounds. 

3. A 2-pound weight and a 3-pound weight on a rough hori- 
zontal table are connected by a thin inextensible cord and moved 
by a constant pull of 7 pounds in the direction of the cord. // = J. 
From the work done by the pull and by friction, find the speed of 
the weights when they have moved 3 feet 9 inches from rest. 

Ans. v = 16. 

4. Show that when two bodies move with the same speed, their 
kinetic energies are proportional to their masses, and that when 
they are acted upon by the same force during the same time their 
kinetic energies are proportional to their speeds. 

5. A weight of W ± pounds rests on a rough horizontal table 
I feet from the edge, and is attached to one end of a light inex- 
tensible cord which passes over a smooth pulley at the edge of the 
table and supports a weight of W 2 pounds hanging h feet above 
the floor. From the work done by gravity and by friction, find 
the speed of the weights when the falling weight reaches the floor, 
and the speed with which the sliding weight reaches the edge of 
the table. 

An, ^ = 2 ^g;^7'> ; «,'=»,'-*>'(»-*). 

6. Show that if, in example 5, 1^ = 10, W 2 = 15 and // = f, in 
order that W t may just reach the edge of table I must be f/i. 

7. Show that the work done by gravity when a body weighing 
W pounds falls to the earth from an infinite distance is Wa foot- 
pounds, the radius of the earth being a feet, and that therefore 
the body will be going V2ga f/s when it strikes the earth. 



Momentum, Work and Energy. 323 

8. Two weights, of W x pounds and W 2 pounds, hang from the 
ends of a light inextensible string which passes over a smooth 
pulley. After t seconds, the heavier weight, W 2 , has descended 
s feet and is going v f/s. Show that 

and by differentiating with respect to t, that the acceleration of 
the weights is 

dv^ W 2 -W, 

dt ~ W 2 + Wj' 

308. Power. — The power exerted by an engine is measured by 
its rate of doing work. The unit of power in the English Gravi- 
tational System is thus the foot-pound per second. 550 foot- 
pounds per second or 33,000 foot pounds per minute make one 
horse-power (H. P.). 

Any unit of force combined with the corresponding unit of dis- 
tance may be used as a unit of work, and with the unit of time 
will form a unit of power. We thus have foot-poundal and foot- 
poundal per second in the English Absolute System, gram-centi- 
meter and gram-centimeter per second in the Metric Gravita- 
tional System. In the Metric Absolute System, the dyne-centi- 
meter unit of work is called an erg, and 10,000,000 ergs a second 
constitute 1 watt. 

If a force of F pounds moves a body weighing W pounds at a 
tangential acceleration of a f/s 2 against resistances having a 
tangential component of R pounds, we have, if the tangential 
component of F is F cos <£ = /, 

f-R=—a > or f=B + W— . 

9 9 

The work done by the force F (or by /) as its point of applica- 
tion moves over any arc s of the path is 



M =i }ds =\i( R+w j) ds - 



324 The Calculus. 



The rate of work is 

Powers d (™f k > = rf ( w f k ) * =v (R+W-0-). 

at as at \ g J 

If the body is being hauled up an inclined plane, making an 
angle with the horizontal, the part of the resistance due to the 
force of gravity is W sin pounds, or the weight times the grade; 
frictional resistances are also proportional to the weight. If the 
total resistance of R pounds is proportional to the weight, and is 
given as h pounds for each pound weight, the source of power 
causing the motion develops 

R + W ~ ) = Wv (fc+— ) foot-pounds per second, 

or 

Wv 



(* + f) 



550 H P 

309. For instance : At the foot of a 3 per cent grade, a 200-ton 
train is going 45 m/h, and at the top of the grade, which is 1 
mile long, is going 30 m/h. If the pull of the locomotive is 
constant and the traction and atmospheric resistances amount to 
20 pounds to the ton, what horse-power does the engine develop? 

20 3 

The total resistance per pound is 9<Mn pound plus the r^ pound 

due to gravity. As the forces are all constant, a is constant, and 
the mile is covered (at a mean speed of 37-J m/h) in — 

seconds, with an increase in speed of —15 m/h= —22 f/s; there- 

- -22x75 11 ., 2 ' , a 11 „ 

fore a— -= — -. _. = — 7B f/ s > an( i — = — tb — ™ • Hence 
2x60x60 48 ' ' g 48x32 

.„. p _ 200x2240 / 3 20 11 \ 

•~ 550 ~ M 100 + 2240 48X32^ 

tj t> 8539 

At the foot of the grade, H. P. = 1164.4 ; at the top, H. P. = 776.3. 



Momentum, Work and Energy. 325 

Again : The driving wheel of a locomotive is 8 feet in diameter, 
the pistons are 2 feet in diameter, the stroke is 5 feet and the mean 
steam pressure in the cylinders is 250 pounds per square inch. 
What EL P. is furnished by the steam pressure in two such cylin- 
ders when the locomotive is going 45 m/h ? 

The driving wheel makes s __„_ revolutions a second. The 

6 8ttX3600 

mean steam pressure is 36,0007r pounds, and in a complete revolu- 
tion, the work done is 360,000tt foot-pounds for each cylinder. 
Hence 

If the engine of a locomotive works under a constant steam- 
pressure, the horse-power it develops is evidently proportional to 

JVv f a \ 

the speed of the locomotive, so that -fkTtI k-\ J —cv, where c 

is a constant. Consequently the acceleration a is constant, and the 
pull of the locomotive is constant. 

If work is to be done continuously by a moving mass, either the 
speed of the moving body must be kept up by the continuous 
application of force, or else the mass must be renewed as fast as 
its kinetic energy is used up. Power is furnished in the first way 
by the fly-wheel of a stationary engine, in the second by a stream 
of water. 

310. Examples. 

1. At what uniform speed can an engine of 30 H. P. draw a 
train weighing 50 tons up a grade of 1' in 280' and against trac- 
tion resistances of 7 pounds to the ton ? Ans. 15 m/h. 

2. An engine under constant steam pressure brings a 500-ton 
train from rest to a speed of 60 m/h in 2 miles against resistances 
of 11 pounds to the ton. What is its H. P. when it has gone 220 
yards, and when it has reached its highest speed ? 

Ans. 733J, 2933f 



326 The Calculus. 

3. A pipe is delivering 11 cubic feet of water a second at the 
rate of 80 f/s; what horse-power is used if the speed is reduced 
one-half? Ans. 93.75 H. P. 

311. Characteristics of Motion. — With the usual notation of 
m, v and f, with L for impulse, K for work, E for kinetic energy, 
P for power, the zero subscript to indicate initial values, and the 
subscript t to indicate tangential components, we have the follow- 
ing relations between the characteristics of motion : 

M=mv, E=imv 2 =iMv, 

L=M-M , K=E-E , 

*~ dt ~ dt ' dt - dt Ji 

The relations on the left involve direction and line of action as 
well as magnitude ; those on the right involve merely magnitude 
and sign. K is the work done by / upon the moving mass; the 
work done by the moving mass against /""is —K = E — E. The 
power of the moving mass is —P. 



CHAPTEE XIV. 

Eigid Bodies. 

312. Mass of a Body of Variable Density. — If any point P of 
a body is surrounded by a closed surface containing a volume AF 

of which the mass is Am, the ratio -r-y. is called the mean density 

of the body within the chosen surface. If the mean density is 
independent of the surface drawn and of the position of P, the 
body is said to be of uniform density, or to be homogeneous. The 
density of the body at any point P is defined as the limit of the 
mean density as the chosen surf ace, always enclosing P, contracts, 
and AT 7 approaches zero. 

Density = p — -^ • 

Then if the density p, at every point of any non-homogeneous 
body is a given function of the position of the point, the mass of 
the body is m — \pdY taken throughout the body. 

For instance, suppose the density of a sphere of radius a is 1 at 
the center and decreases by an amount proportional to the distance 
from the center, becoming f at the surface. The density at a 
distance r from the center is p = l — At, and as p = f when r = a, 

k = -^— , and p = a ~ . A spherical shell of radius r and thick- 
3a 3a 

ness dr therefore has for its mass dm— ^— r 2 (3a — r)dr, approxi- 

mately, and the total mass is 

m=p- \ a (3ar 2 -r 3 )dr = Tra 3 . 
3a Jo ' 



328 



The Calculus. 



313. Resultant of Like Parallel Forces. — We have so far con- 
sidered only such forces as have concurrent lines of action. Two 
forces of which the lines of action are parallel are called like or 
opposite parallel forces, according as their directions are the same 
or opposite. 

The resultant of two parallel forces can be found by combining 
the triangle construction with the principle that the point of 
application of a force acting on a rigid body can be -shifted along 
the line of action without altering the effect of the force. 



*U-- 




Fig. 104. 



Let P and Q (Fig. 104) be two like parallel forces, A and B 
their respective points of application. Join AB, and let equal 
and opposite forces —M and M be applied at A and B, having AB 
as their common line of action. These two forces neutralize each 
other, so that, R being the resultant of —M and P, and 8 the 
resultant of M and Q, the pair of forces R and 8 is equivalent to 
the given pair, P and Q. In order to find the resultant of R and 
8, shift their points of application to the common point of 
their lines of action, and resolve each of them in the direction of 
AB and that of the forces P and Q. Evidently the components 
of R are — M and P, and of 8 are M and Q, so that the resultant 



AC 
OC 


M 

~ P > 


BC 

OC 


M 
" ' 


AC 
BC 


Q 

" P ' 







Eigid Bodies. 329 

of R and #, which is also the resultant of the forces P and Q in 
their original positions, is a force T, of magnitude T—P-^-Q, 
having the same direction as P and Q. Moreover, if T's line of 
action meets AB at C, we have from similar triangles, 



or 



The magnitude, the direction and the line of action of the 
resultant of the given forces P and Q are thus completely deter- 
mined. 

If a common perpendicular to the lines of action of P , T and 
Q is divided by them into segments a and b, 

¥ = W = -p> or Pa = Qh - 

314. Moment of a Force. — The moment of a force about a 
straight line perpendicular to its line of action is defined as the 
product of the magnitude of the force by the length of the com- 
mon perpendicular to the line of action and the given straight 
line if the rotation that the force tends to produce about the 
straight line as an axis is contra-clockwise, and as the negative of 
this product if the rotation is clockwise. 

When the forces considered all lie in the same plane, so that the 
straight lines about which moments are taken (the axes of the 
moments) are all perpendicular to this plane, it is convenient to 
speak of the moment of a force about an axis as its moment about 
the point in which the axis cuts the plane. 

315. Moments of Parallel Forces. — The moments of the forces 
P and Q of Art. 313 about a line perpendicular to their plane at 
any point of the line OC (about any point of OC) are Pa and 



330 The Calculus. 

— Qb; the moment of T about the same axis is zero, or the sum 
of the moments of P and Q. 

Consider the moments of P, Q and T about any point 0' of 
their plane. For convenience, shift the forces along their lines 
of action until their points of application, A, B and 0, fall on 
the perpendicular to their lines of action through 0' (Fig. 105). 
Then AC=a, OB=b; call 0'A=x. 

The moments about 0' are : Of P : —Px; ofQ: — Q(a+b + x); 



31 jc ^ &__/?_ A A. 



T 



Fig. 105. 



of T = P + Q: -(P + Q)( a + x). The sum of the moments of 
P and Q, since Qb—Pa, is 

- ^Px+Qa+Qb + Qx] = - {P + Q) (a + x), 

or the moment of T. The same result is obtained if 0' is between 
A and B or on the other side of B. 

Consequently, the resultant of two like parallel forces of mag- 
nitudes P and Q is a third like parallel force of magnitude 
(P+Q), having for its moment about any axis perpendicular to 
the common plane the sum of the moments of the two given forces 
about the same axis. 

It evidently follows that the resultant of any number of like 
parallel forces is a force having the same direction as its com- 
ponents, a magnitude equal to the sum of their magnitudes, and 
a line of action such that its moment about any axis perpendicular 
to the common direction of the forces is the sum of the moments 
of all the forces about the same axis. 



Rigid Bodies. 331 

316. Resultant of Opposite Parallel Forces. — The process of 
Art. 313, if applied to a pair of opposite parallel forces of different 
magnitudes, shows that their resultant has the same direction, and 
tends to produce rotation in the same direction, as the larger 
force ; that its magnitude is the difference between the magnitudes 
of the two forces, and that its line of action is so situated in the 
plane of the two forces that the moment of the resultant about 
any point of the plane is the difference between the moments of 
the two given forces about the same point. 

Consequently, the results stated in Art. 315 for like parallel 
forces will hold with one exception for any parallel forces if the 
magnitudes of forces in one of the two directions are represented 
by positive numbers, and those of forces in the opposite direction 
by negative numbers. 

317. Couples. — The exception occurs when the forces reduce to 
two opposite parallel forces of the same magnitude, with different 
lines of action. Such a pair of forces, called a couple, can be 
replaced or balanced only by another couple. The sum of the 
moments of the forces constituting the couple, called the moment 
of the couple, is of course even in this case the same for any axis 
as the sum of the moments of the original forces ; aside from this 
fact, we shall not for the present be concerned with the nature of 
couples. 

318. Identical and Identically Opposed Forces. — Two forces, 
f x and f 2 , having the same magnitude, the same line of action and 
the same direction, are identical ; this is indicated by writing : 
f x = f 2 . Two forces f x and / 2 , having the same magnitude, the 
same line of action, and opposite directions, are said to be equal 
and directly opposed, or balanced; this is indicated bv writing: 

f,--7«- 

Note that fi — f 2 or f x ——f 2 imply magnitude and direction, 
but not line of action. 



332 The Calculus. 

Two forces produce no effect on the motion of a body when and 
only when they are equal and directly opposed; in other words, 
if a body is at rest under the action of parallel forces, the algebraic 
sum of all the forces is zero, and the sum of the moments of all the 
forces about any axis is zero, for the resultant of the forces having 
one of the two directions must be equal and directly opposed to 
the resultant of the forces having the opposite direction. 

319. Balanced Parallel Forces. — The lever furnishes the sim-. 

plest instance of balanced parallel forces ; the following problem 

is of the same nature. 

A light circular table, 10 feet in diameter, is supported on three 

vertical legs, spaced at equal angular intervals, each 4 feet from the 
center. Let be the center of the table-top, 
A, B, C the points directly over the legs. It 
is required to find the tensions or compres- 
sions in the legs if the table is stationary 
and of negligible weight and supports a 
weight of 60 pounds placed on the line OA : 
(1) on the edge of the table at the point 

nearest A, (2) at the point farthest from A, 
Fig. 106. ^ between B and ^ ^ ftt ^ (5) flt A 

Consider the forces acting on the table- top ; these are the weight 
of 60 pounds and the pressures or pulls of the legs, which are di- 
rectly opposite to the compressions or tensions in the legs. Let A 
represent the pressure exerted by the leg at A (or the compression 
in A), and so for the other legs; then negative values will indi- 
cate pulls exerted by the legs, or tensions in the legs. 

In any of the cases, we have 

A+£+C=60, 

and if we take moments about the line AO y we find B = C. Tak- 
ing moments about the perpendicular to AO at we find ikf + 4A 
— 2(B + C)=0, where M is the moment of the 60-pound weight, 




Rigid Bodies. 333 

and is —300, 300, 120, and —240 in the several cases. Solving 
the three equations in each case we find : 



Case 


1 


2 


3 


4 


5 




70 


-30 





20 


60 


= C = 


-5 


45 


30 


20 






320. Examples. 

1. Show that in the case of any lever, if we neglect the weight 
of the lever, the work done by the power in any motion is numeric- 
ally equal to that done by the resistance. 

2. A triangular table, sides 6 feet, 8 feet and 10 feet, has a 
vertical leg under the middle point of each side. Neglecting the 
weight of the table, find the compressions in the legs when a 
weight of 300 pounds rests at a point 3 feet from each of the 
perpendicular sides. 

Ans. 75 pounds, and 225 pounds respectively. 

3. Find the compressions if in example 2 the weight is placed 
2 feet from each of the perpendicular sides. 

Ans. 150 pounds, 100 pounds and 50 pounds respectively. 

4. Solve example 3 with the legs at the corners. 

Ans. 125 pounds at the right angle, 75 pounds at the smaller 
acute angle, 100 pounds at the larger. 

321. Center of Gravity of a Body. — Suppose three mutually 
perpendicular coordinate planes to be fixed in a material body 
and the body to be placed so that the axis of z is vertical. Let the 
body be divided up into elements of volume, of which dV is 
typical, and let the density of the body at a point of this element 
be p, so that the mass of the element is p • dV ' — dm. 

Then the body may be conceived as made up of heavy particles 
of which the element of mass dm is a type, and the force of 
gravity acting on the body (the weight of the body) may be 
regarded as the resultant of a system of like parallel forces, the 
forces with which gravity acts on the constituent particles, of 
which gdm is a type. When these forces are summed by integra- 
tion, the approximate assumptions become exact; consequently, 



334 The Calculus. 

the force with which gravity acts on the body has for its magni- 
tude the integral of gdm taken throughout the body, or 
g\dm — gm, m being the mass of the body, and gm — W, its weight. 
Suppose the line of action of the resultant force of gravity (which 
is vertical) to be at the intersection of the planes x = x , y = y 0? 
and consider the moments of the resultant about the axes of x 
and y, each of which is perpendicular to all the forces. The 
moments of an element gdm, situated at a point (x, y, z) of the 
body are ygdm about the axis of x and xgdm about the axis of y. 
The sum of all the elementary moments about either of the axes 
is the moment of the resultant, W—mg, about the same axis; 
hence 

™>gyo=Syg dm ; m 9 x o = fagdm, 

or 

_ \xdm \xdm _ \ydm \ydm 

X °~ ~m~ ~ Jdm~ ' Vo ~ ~~m~~ ~ JdnT ' 

the integrations in each case being taken throughout the body. 

If we suppose the body to be set up with the .T-axis or the y-axis 
vertical, we find y or x as before, and 

_ \zdm _ \zdm 

m ~~ \dm 

The point {x , y , z ), through which the resultant weight of 
the body always acts, is called the center of gravity of the body. 
Each of its coordinates is the mean distance of points of the body 
from the corresponding coordinate plane, the distribution being 
proportional to the mass, or to both the volume and the density. 

In the case of a homogeneous body, the factor p in dm=pdV 
is constant, so that 

_ pjxdV _ SxdV _ SydV _ \zdY 

The center of gravity in this case is called the center of gravity 
of the geometric solid occupied by the body. 



Eigid Bodies. 335 

322. Centers of Gravity of Areas and Arcs. — It is often im- 
portant to find the center of gravity of a geometric surface or arc ; 
in such problems the element of volume dV is replaced by the 
element of surface dS, or by the element of arc ds. 

The coordinates of the center of gravity of a plane area, if dA is 

the element of area, are thus given by the integrals : x = f AA , 
y — f-j A > taken over the area ; and for an arc of a plane curve, 

the coordinates of the center of gravity are given by the integrals : 

x = J^ ,y =Syte taken along the arc. 
)ds )ds b 

It is only in the case of a plane surface that the center of 
gravity lies on the surface, and except in the case of a straight 
line, the center of gravity of an arc of a plane curve never lies on 
the curve. 

If a body is in any way symmetrical, the location of its center 
of gravity is always partly evident; for instance, the center of 
gravity of any homogeneous solid of revolution is evidently on 
the axis of revolution, and the center of gravity of any homo- 
geneous central figure is at the geometric center. 

323. Work Done by Gravity on an Extended Body. — If an 

extended body weighing W pounds moves so that its center of 
gravity falls h feet the work done by gravity is + Wh foot-pounds, 
for the resultant force of gravity is applied at the center of 
gravity, which moves in the direction of the force. In moving 
the body so that its center of gravity rises h feet, Wh foot-pounds 
of work are done against gravity. For instance, to up-end a 
24-foot ladder weighing 50 pounds requires 600 foot-pounds of 
work, and to pump out a cistern 10 feet deep containing 120 
cubic feet of water standing 4 feet deep, requires 62.5x120x8 
f= 60,000 foot-pounds of work. 



336 



The Calculus. 



324. Computation of the Coordinates of Centers of Gravity.— 

As a coordinate of the center of gravity of a body is merely the 
mean distance of the points of a body from the corresponding 
plane, there is nothing new about the process of computing the 
coordinate by an integration. In the first illustrative example of 
Art. 231, and in examples 4a and 10 of Art. 233, we found the 
distance of the center of gravity from the base in the case of the 
semi-circumference, the hemispherical surface and the hemisphere 



of radius a to be 



2a 



3a 



2 and -£- respectively. In the case of 



4a 



the semi-circle, the corresponding distance is ^— . 

The center of gravity of a triangle is between each vertex and 
the mid-point of the opposite side, twice as far from the vertex 
as from the opposite side; the center of gravity of any pyramid 
or cone is between the vertex and the center of gravity of the base, 
three times as far from the vertex as from the center of gravity of 
the base. 

Bearing in mind that the center of gravity is the point at which 
the resultant force of gravity acts, we can use these elementary 
results for the purpose of finding certain centers of gravity without 
integration. 

For instance, let it be required to find the center of gravity 
t of the trapezoidal area in Fig. 107. Divide 
the figure into a triangle and a rectangle, as 
shown. Let area = weight for convenience 

f p = — ) , and consider the moments of the 

9' triangle and the rectangle about the 6' and 

4' 6" sides. The sums of these moments are 

the corresponding moments of the whole area. 

Then if x and y are the distances of the re- 

4'&" -* quired center of gravity from the 6' and 4' 6" 

Fig. 107. sides, respectively, 







Eigid Bodies. 337 

46Xf+iX3X|] = (iX|)(6X|) + (|Xf)(iX|X3), 
?y[6Xf+i-X3X|]=(|X6)(6X|) + (6+|X3)(ixfX3), 

whence x =2A f , y = 3.8' 

Again, to find the center of gravity of a solid consisting of a 
hemisphere and a cone having a common base, 
the vertical angle of the cone being 90°. Let 
the radius of the hemisphere be a; then 
the altitude of the cone is a. Take moments 
about a diameter of the base. [The figure 
shows the section of the solid by the plane 
through the geometric axis and perpendicular 
to the axis of moments.] F ^ 108 

Let volume = weight, and let the distance 
of the required center of gravity from the common base be x. 
Then 

X ( fTra 3 + l-rra? ) = -^- X f Tra 3 - ±. X faa*, 



325. Examples. 

1-3. Find by integration the centers of gravity of the follow- 
ing homogeneous figures (see Art. 324) : 

1. Semi-circle. (Find the moment about the base of an ele- 
ment perpendicular to the base, considering its mass to be con- 
centrated at its mid-point.) 

2. Triangle. 

3. Cone or pyramid. 

4. Show that the solid mentioned in the second illustrative 
example of Art. 324 will stand with any point of the hemispherical 
surface in contact with a horizontal plane if the vertical angle of 
the cone is made 60°. 

5. How much work is done by gravity in emptying a cone full 
of mercury through a hole in its base into a cylindrical beaker 
having the same base as the cone, if the altitude of the cone is 

23 



338 The Calculus. 

1 foot, the radius of its base 6 inches and the base of the cone is 
5 inches above the base of the cylinder? [Take tt— 2 ^, s. g mer- 
cury =14.] Ans. 114 -^2 foot-pounds. 

6. Show that if the table top in the illustrative example of 
Art. 319 weighs 20 pounds, each of the results is increased by 6§. 

7. A chain 20 feet long, weighing 15 pounds to a foot, hangs 
down from the deck of a ship into the hold; what work is done 
against gravity in hauling it up on deck ? In hauling up the first 
10 feet of its length ? 

Ans. 3000 foot-pounds, 2250 foot-pounds. 

8. One end is turned off a cylinder of revolution 9 feet in 
length, so that a solid is left consisting of a cylinder 8 feet in 
length and a cone of altitude 1 foot. Find the center of gravity 
of the solid. 

Ans. 3.83 feet from the common base of the cylinder and the 
cone. 

9. One end is turned off a cylinder of revolution 9 feet in 
length, leaving a cylinder 8 feet in length capped by a hemisphere. 
Find the center of gravity of the solid. 

Ans. 3.66 feet from the common base of the cylinder and the 
hemisphere. 

10. Find the center of gravity of the part of the ellipse, 

x — a cos <j>, y— b sin <f> in the first quadrant. 

A 4a 46 

Ans. x Q = —,y =—. 

11. Find the center of gravity of a circular arc of 90°. 

2\/2 
Ans. a— 0.9003a from the center. 

12. Find the center of gravity of a broken line composed of a 
circular arc of 90° and its chord. 

Ans. ■ ,— a = 0.8088a from the center. 

13. Find the center of gravity of one of the halves into which 
a parabolic segment of altitude a, base 2b, is divided by the 
altitude. 

Ans. fa from the base, f l from the altitude. 



Rigid Bodies. 339 

14. Find the center of gravity of a paraboloid of revolution of 
altitude a, radius of base 0. 

Ans. fa from the vertex. 

15. Find the center of gravity of the solid formed by revolving 
one-half the parabolic segment of example 13 about the base. 

Ans. yg-fr from its base. 

16. Find the center of gravity of the solid left when a pyramid 
is cut from a cube of edge I by a plane passing through a diagonal 
of one face and a vertex of the opposite face. 

Ans. It lies on the severed diagonal of the original cube, 

■jr= V3 from the center of the cube. 

17. Find the center of gravity of the area between one arch of 
the cycloid x=a(cf> — sin<£), y = d(l — cos<f>) and the z-axis. 

Ans. y = §a. 

18. Find the center of gravity of the half above the initial line 

of the area bounded by the cardioid r = 2asin 2 ^ . 



Ans. x n — 



2 

5a 16a 



6 '" 9tt 

19. Find the center of gravity of the sold formed by revolving 
the figure of example 18 about the initial line. 

Ans. x =—£a. 

20. A solid is formed of a hemisphere of radius a and a solid 
of revolution of height h, the two having a common base. What 
must be the value of h for the solid to stand with any point of its 
hemispherical surface in contact with a horizontal plane, the 
solid of revolution being (1) a cylinder, (2) a paraboloid? 

Ans. (1) h=~ V2, (2) fc=JLV6. 

326. The Pappus-Cavalieri Theorems. — The first of the follow- 
ing very useful theorems was contained in a compilation of mathe- 
matical knowledge made by Pappus, of Alexandria, about 300 
A. D. ; it was proved by Cavalieri, an Italian, about 1629, by 
his famous " Method of Indivisibles " ; it was announced at 
about the same time by a G-erman, G-uldin, who appropriated it 
from Pappus. It is commonly known as " Guldin's Theorem." 



340 



The Calculus. 



I. If a plane area is revolved about an axis in its own plane 
which does not pass through it, the solid generated has a volume 
equal to the product of the given area by the length of the path 
traced by its center of gravity. 

II. If an arc of a plane curve is revolved about an axis in its 
own plane which does not pass through it, the surface generated 
has an area equal to the product of the length of the arc by the 
length of the path traced by its center of gravity. 

In each case, let x Q be the distance of the center of gravity G 




Pig. 109. 



from the axis of rotation, and let x be the distance of any point P 
from the same axis. In I, let dA be the element of area at P, 
then if A is the revolved area, and V the volume it generates when 
its plane is turned through the angle a, 



and 



A = $dA, Y—\xadA—a\xdA, 
\xdA 



x n — 



SdA 



All these integrals are taken over the same area; hence 
V = Ax a, which proves I, since x a is the length of the path 
traced by G. 



Eigid Bodies. 341 

In II, let ds be the element of arc at P; then if s is the length 
of the revolved arc, and S the surface it generates when its plane 
is turned through the angle a, 

s—\ds, 8=\xads — a\xds, 
and 

\xds 

All these integrals are taken over the same arc ; hence 8=s- x a, 
which proves II, since x a is the length of the path traced by G. 

For instance, if a circle of radius a is revolved about a tangent 
through 2ir, its center describes a path 2-n-a in length, and the 
volume and surface of the resulting solid are TTa 2 x2-na—2ira?' 
and 27rax27ra = 4:Tr 2 a 2 . 



327. Examples. 

1. A rectangle of which the sides are 1 foot and 1 foot 8 inches 
in length is revolved about an axis in its plane parallel to the 
1-foot sides and 1 foot 6 inches from the nearer one. Find the 
volume of the ring so formed. 

•22 

Ans. 24-f- cubic feet, using tt— '-=- . 

2. Given that the center of gravity of a half parabolic segment 
is fa from the base, §& from the altitude (example 13, Art. 325), 
find the volumes formed by revolving this area about the base and 
about the altitude. 

Ans. T 8 5-7ra 2 &, %irab 2 . 

3. From the known values of the surface and volume of a 
sphere find the positions of the centers of gravity of the arc and 
the area of a semi-circle. 

4. A semi-circle is revolved about a tangent at its vertex ; find 
the volume and the surface of the solid generated. 

Ans. a 3 (7r 2 -f7r)= 5.6808a 3 ; inner surface 27ra 2 (7r-2) = 
7.1728a 2 , outer surface 4rra 2 . 

5. From the result of example 4, determine the center of 
gravity of the area enclosed by a circular arc and two perpendicu- 



34-2 The Calculus. 

lar tangents, and find the volume of the solid generated as this 
area revolves about the chord of the quadrantal arc. 

-j a q 

Ans. — — r- a = 0.2234a from each tangent, volume = 

o(4 — tt) 

T V7rV2(37r-8)a. 3 = 0.5275a 3 . 

6. From the results of examples 8, 12, 13, Art. 191, locate the 

center of gravity of an arch of the cycloid and that of the upper 

half of the arc of the cardioid. 

Ans. Cycloid, y =±a; cardioid, x = — §a, y — ^a. 

328. Wind Pressure on a Plane Surface. — If the pressure 
exerted by the wind on a plane surface normal to the direction of 
the wind is uniform over the surface, and amounts to p pounds 
per square foot, the wind pressure on a plane surface whose 
normals make the angle 9 with the direction of the wind will be 
p cos pounds per square foot. The point where the line of action 
of the resultant wind-pressure meets the plane is called the center 
of effort of the pressure, or the center of wind-pressure. The 
process of finding this center of effort is precisely the same as that 
of finding the center of gravity of the area under pressure, for 
the forces acting are in both cases proportional to the area over 
which they act, the elements of pressure being p cos OdA in one 
case, ffpdA in the other. Consequently, the center of wind-press- 
ure for a plane area is the same as the center of gravity. 

The center of wind-pressure for a sail is determined by the 
method exemplified in the example of the trapezoidal area in 
Art. 324. 

329. Total Fluid Pressure and Center of Fluid Pressure for a 
Plane Surface. — The pressure in a fluid is found to be the same 
in all directions at any one point, and to be exerted against any 
surface in a direction normal to the surface. In a fluid having 
no vertical motion, the upward pressure on the base of a vertical 
column h feet high and having a cross-section of 1 square foot, is 
therefore the weight of h cubic feet of the fluid plus whatever 



Eigid Bodies. 343 

pressure there is on 1 square foot of the upper surface. If the 
fluid weighs w pounds to the cubic foot, the pressure due to the 
weight of the fluid alone is thus wn pounds per square foot at a 
depth of h feet. Fresh water weighs 62^ pounds per cubic foot, 
sea water 64 pounds. At a depth of h feet in fresh water there 

is a pressure in every direction of f p -| — ) pounds, the 

pressure on the upper surface being p pounds per square foot. 
In an open body of water, p is the atmospheric pressure, 15 
pounds per square inch when the barometer stands at about 30^ 
inches. In many cases, the atmospheric pressure acts on a sur- 
face equally in opposite directions, so that only the pressure due 
to the weight of water, called the water-pressure, need be con- 
sidered. 

330. Water-Pressure on a Vertical Plane. — Suppose a plane 
surface to be submerged vertically in still water ; to find the total 
pressure on it, due to the weight of the water (called the total 
water-pressure) , and the line of action of the resultant pressure 
(intersecting the plane of the surface in a point called the center 
of pressure). 

Take as the axis of x the horizontal line "in which the plane of 
the submerged surface cuts the surface of 
the water, and as the axis of y any con- — 
venient vertical line in the plane of the sub- 
merged surface. Divide the surface into in- 
finitesimal elements of which dA at the point 
(x, y) is typical. Then the water-pressure 
on the element dA is ivydA in magnitude, y 
and the resultant of all such elementary F 110 

pressures has for its magnitude \wydA 
taken over the submerged surface, for the elementary pressures 
are parallel forces. 




344 The Calculus. 

If the line of action of the total water-pressure intersects the 
plane of the submerged surface at (x , y ), the moments of the 
total pressure about the axes are x $wydA about the y-axis, and 
y Q \wydA about the #-axis. The moments of the elementary 
pressure wydA at (x, y) are wy 2 dA about the #-axis, and wxydA 
about the 2/-axis. Hence 

x \wydA — \wyxdA, y^wydA — \wy 2 dA, 
or 

_ \wxydA _ \wy 2 dA 

X °~ SwydA ' y °~ $wydA ' 

331. If (x r , y ') is the center of gravity of the submerged 

vertical plane surface considered homogeneous,, y '= ?-, A — 

— , A being the area of the submerged surface, and the total 



A 

water-pressure, 

P = j wydA = w\ydA — wy 'A ; 

that is, the total pressure on a submerged vertical plane surface 
is the same as if the surface were horizontal and at a depth equal 
to the actual depth of its center of gravity; in other words, the 
mean water-pressure on the surface is the pressure at the center 
of gravity. 

332. Water-Pressure on an Inclined Plane. — The total water- 
pressure on an inclined plane surface and the center of pressure 
can be found in essentially the same way by taking the axis of x 
as before and the axis of y in the inclined surface. If the in- 
clination of the plane of the surface to the vertical is a, a point 
(x, y) of the surface is at a depth y cos a, so that the total water- 
pressure is 

P — ^wy cos adA — wy r cos a • A, 



Rigid Bodies. 



345 



and is again the submerged area multiplied by the pressure at its 
center of gravity, y ' being the dis- 
tance of the center of gravity from the 
#-axis, measured on the inclined plane. 
Again, (x , y ) being the coordinates 
(measured on the inclined plane) of 
the center of pressure, the equality of 
the moments about the coordinate axes 
of the total pressure and the sum of 
the elementary pressures gives 

x Q \wy cos ad A — \wxy cos ad A, 
y^wy cos ad A — \wy 2 cos ad A, 

whence r and y have the same values 
independently of a. 




Fig. 111. 



333. Computation of Centers of Pressure. — From Arts. 331 
and 332 it appears that the center of pressure for a plane area is 
the same point of the area, however the plane may be revolved 
about its intersection with the free water-surface; and the total 
pressure is unchanged by any revolution of the area about a hori- 
zontal axis through its center of gravity. 

The preceding discussion is typical of the process of finding 
centers of pressure, but the details of the method can often be 
altered to advantage. For convenience, we always consider the 
plane of the submerged area to be vertical, and take one of the 
axes, say the #-axis, horizontal; but the origin should be chosen 
so as to simplify as much as possible the relations between y and 
x on the boundary of the submerged area. For instance, if this 
area is a triangle, the origin should be taken at a vertex ; if it is a 
circle, at the center. 

In such a case, if the depth of the origin below the free sur- 
face is c, and that of the center of gravity of the area A is h, the 
total pressure P and the coordinates of the center of pressure 



346 . The Calculus. 

are given by 

P = whA J Px = $wx(y + c)dA, Py = $wy(y + c)dA; 

and the depth of the center of pressure is (y + c). 

If the plane of the submerged area is inclined at the angle a 
to the vertical, we have only to change the last result to 
(y + c) cos a. 

In evaluating y 0> it is always possible to take as the element of 
integration a strip of the area between two horizontal lines dy 
apart, and when rectangular coordinates are used, it is best to 
do so. If the area is symmetrical with regard to a vertical line, 
the center of pressure lies on that line; otherwise, x may be 
evaluated by a double integration, or by using the horizontal 
strip as the element of integration and replacing x in the formula 
by the abscissa of the center of gravity of the element. 

334. Examples. 

Find the total fluid pressure and the depth below the free sur- 
face of its center of effort for each of the following vertical plane 
areas : 

1. Kectangle; breadth ~b, height h; upper edge in surface. 

Ans. P=iwbh 2 , y = %h. 

2. Triangle; base o } horizontal; altitude h; vertex in surface. 
Ans. P=iwbh 2 , y = %h, c. p. on median through upper 

vertex. 

3. Triangle; base b, in surface; altitude h. 
Ans. P—\mbW-, y =iK c. p. on median through lower vertex. 

4. Quadrant of circle ; radius a, one edge in surface. 

Ans. P=iwa?, y = — | a, x = %a. 

5. Circle; radius a, just submerged. 

Ans. P=7rwa?, y =%a. 

6. Ellipse ; semi-axis a vertical, center in surface. 

Ans. P=%wa?b, yo=jjr*> 



Rigid Bodies. 347 

7. Parabolic segment; base %b, in surface; altitude h. 

Ans. P=^wh 2 b, y = jh. 

8. Isosceles trapezoid; base 3 feet in surface, base 5 feet at a 
depth of 4 feet. 

Ans. P = 2166f pounds, */ =fffeet=2feet 9.23 inches. 

Find the total fluid pressure and the depth D of the center of 
pressure below the center of gravity for each of the following 
completely submerged vertical plane areas, h being the depth of 
the center of gravity below the free surface. 

a 2 

9. Square ; side a. Ans. P = iva 2 h, D— -^-. 

a 2 

10. Circle; radius a. Ans. P = Trwa 2 li, D = — ri -. 

11. Triangle; base %b, horizontal; altitude a; vertex up. 

Ans. P = tuabh, ^=ToTT* 

12. Parabolic segment; base 2b, horizontal; altitude a; vertex 
up. 

Ans. P=±wabh, D=^-. 
3 175/i 

335. Kinetic Energy of an Extended Body. — If a material 
hody moves so that all its points have the same velocity — a speed 
of v f/s in a given direction — the kinetic energy of an element 
dm of its mass is \v 2 dm, and the kinetic energy of the whole body 
is \\v 2 dm—\v 2 \dm—\mv 2 . 

The motion just described is called translation; it is only in 
translation, when the paths of motion of the points of the body 
are parallel straight lines, and the body is rigid, that the kinetic 
energy is \mv 2 . 

Any motion of an extended body is the resultant of a rotation 
about an axis (which may itself be moving) and a translation 
along the axis ; the kinetic energy of a body in any motion is the 
sum of , the energies due to these two motions. 



348 The Calculus. 

336. Kinetic Energy of a Rotating Body. — Suppose a body to 
be rotating about a fixed axis at the angular rate of <o radians a 
second. Imagine the body divided up into infinitesimal elements 
of volume by some three sets of surfaces, and consider each of the 
divisions of the body as a heavy particle. Let the volume of any 
one of these particles be dV, and the density p, so that its mass is 
dm—pdY, and let its distance from the axis of rotation be r feet. 
The particle is moving in a circle of radius r feet with an angular 
speed of w radians a second, or with a speed in its path of v = o>r 
f/s. Its kinetic energy is therefore dE —\dmv % —\dmh?r 2 '. The 
kinetic energy of the whole body is the sum of the kinetic energies 
of all the particles, or is 

E = jJwVdra =i« 2 jr*dm, 
the integration being taken throughout the body. 

If the body is homogeneous, p is constant, and E=i<o 2 p$r 2 dV. 

337. Moment of Inertia and Radius of Gyration. — The quan- 
tity \r 2 dm, which bears the same relation to the angular speed 
and kinetic energy of a rotating body that the mass bears to the 
linear speed and kinetic energy of a body in translation, is called 
the body's moment of inertia with reference to the given axis, 
and is indicated by /. 

The value H , = — is the mean value of the squared dis- 
)dm m 

tance from the axis of rotation of points of the body, the distri- 
bution being proportional to the mass, or to both volume and 
density. This mean value is indicated by Tc 2 ; as l—mk 2 , k is the 
distance from the axis at which a heavy particle of the same mass 
as the body would have the same moment of inertia, or the same 
kinetic energy when rotating about the given axis at the same 
angular rate, h is called the radius of gyration of the body with 
reference to the given axis. 

338. Computation of I and k 2 . — Moments of inertia are of 
fundamental importance, not only in the consideration of rotat- 
ing bodies, but also in many other connections, such as the bending 



Eigid Bodies. 349 

of beams and the stability of ships. The moments of inertia of 
homogeneous areas are used in the latter problems, and are con- 
venient in the computation of moments of inertia of material 
bodies. 

If p is the density of a body at an element of volume dV ' , the 
element of mass is dm — pdV, and 

i=^ P dv ; y=-L = Sft£r. 

J r m ) p dV 

If the body is homogeneous, p is constant, and 

L-p)rav , ft - m - ^ dy --^- . 

The transition to the conceptions of the moment of inertia 
and squared radius of gyration of an area or an arc, with dA or 
ds in place of dV, is the same as the corresponding process in the 
case of centers of gravity. In what follows, the density will be 
supposed uniform if not specified. 

A few simple moments of inertia and certain general principles 
aid materially in the actual computations. It is a part of the 
definition that the same moment of inertia is obtained from a 
given mass at a given distance from a given axis whether the mass 
is concentrated at a point or distributed in any way over the 
circumference of a circle or the surface of a cylinder, and that the 
moment of inertia of a system of particles or bodies is the sum of 
the moments of inertia of its constituent parts. 

Thus, for a circumference of radius a with reference to an axis 
perpendicular to its plane through its center, or for a cylindrical 
surface of revolution of radius a with reference to its geometric 
axis, h — a. 

For a rectangle with reference to an axis in its plane parallel 
to one of its sides, the radius of gyration is the same as for 
one of the perpendicular sides; and for any right cylinder with 
reference to an axis parallel to its geometric axis, the radius of 
gyration is the same as for a section normal to its axis. 



350 



The Calculus. 




For, let the dimensions of the rectangle be I and b, and let the 
axis be parallel to b (Fig. 112). Then if lines 
are drawn across the rectangle parallel to the 
axis, the moment of inertia and the mass of the 
rectangle are the same as if the mass of every 
such line were concentrated at the point where 
the line crosses I, and so are what the correspond- 
ing values for the line I would become if the 
density of the line were b times that of the rec- 
tangle. As the radius of gyration is independent of this density,, 
the first part of our proposition is proved. , 

The second part of the proposition is proved in the same way 
by drawing cylindrical surfaces of revolution having the axis of 
reference as a common axis, and supposing the mass of every such 



Fig. 112. 




Fig. 113. 



surface to be concentrated along the arc in which the surface cuts 
the normal section of the given cylinder. 



339. Examples. 

Find the values of I and h 2 for each of the following homo- 
geneous figures, the first five by direct integration, the rest by 
combining values already found. 

1. Straight line of length I; axis perpendicular to the line at 

I 3 I 2 

one end. Ans. l—-— i 'k 2 —-~. 

o o 

2. Straight line of length I; axis perpendicular to the line 
extended, and distant c from the nearer end. 

Ans. I = cl(c + l) + |-, fc 2 = c(c+Z)+ | . 



Eigid Bodies. 



351 



3. Eectangle of length I and breadth b ; axis one of the sides of 

l 3 b I 3 

length I. Ans. 1= -~- , lc 2 —-—. 

4. Eectangle of length I and breadth b; axis parallel to the 
sides of length b and distant c from the nearer one. 

Ans. I = clb(c + l)+^-,k 2 = c(c + l)+^. 

5. Triangle of altitude h and base b ; axis parallel to the base 
and through the opposite vertex. 

ih 2 . 











Ans. 


1 = 


\bh\ 


Tc 2 = 




, 


' 


?" 




/'6" 










a" 




6 








3 


3" 3" 




6" 


8" 








F. 


[G. ] 


.14. 




Fig. 115. 



6. The accompanying T-beam section, axis as shown in Fig. 
114. Ans. 1 = 540, fc 2 =15. 

7. The hollow square in the accompanying figure (Fig. 115) ; 
axis shown. Ans. 7 = 31,968, k 2 — lll (inch-units). 

8. Two straight lines, each of length I, perpendicular to each 
other at their mid-points ; axis parallel to one through an end of 
the other. Ans. I—^l B i k 2 =^l 2 . 

9. A pole AB, standing upright with the end B on a horizontal 
plane, falls without sliding. Find the speed with which A hits 
the plane if AB = 2± feet (the kinetic energy is acquired through 
the work done by gravity during the fall) . Ans. 48 f/s. 

340. Perpendicular Axes for a Plane Area. — If Jc x 2 and Jc y 2 
are the squared radii of gyration for a plane area or arc with 
reference to two perpendicular axes lying in its plane, the squared 



352 The Calculus. 

radius of gyration with reference to an axis perpendicular to the 
plane at the intersection of the given axes is 

Let the three mutually perpendicular axes be OX, OY, OZ, 
and take them as axes of coordinates. Then the moments of 
inertia of the given area with reference to these axes are: 

I x —\y 2 dA, I y —\x 2 dA, and I z — \r 2 dA, 

where, since r 2 = x 2 + y 2 , 

I z = \x 2 dA + Sy 2 dA =I V +I X . 

Then 

A. Kg — jljl Kx i~ A- toy i 

or 

Kg =z Kx T" Ky . 




Fig. 116. 



An exactly similar proof establishes the 
proposition in the case of an arc. 



341. Parallel Axes for any Body. — If h a 2 and k g 2 are the 

squared radii of gyration for any body with reference to any axis 
a and an axis g, parallel to a through the center of gravity G; 
and if R is the distance of G from a, then 

Kq, — Kg -J- -ti • 

Choose a set of perpendicular axes, taking g as the axis of z, 
and let a cut the plane of xy at (h, k) (see Fig. 117). 

Then I g — - mk g 2 '■ = \ (x 2 + y 2 ) dm, and I a = mk a 2 = §r 2 dm, where 
^=( x -hy + ( y - K y and h 2 + k 2 = R 2 ; i. e., r 2 = (x 2 + y 2 ) +R 2 
— 2hx — 21cy. 

From the formulas for the coordinates of the center of gravity, 

_ \xdm q_ \ydm 
~ \dm ' ~ \dm ' 



Eigid Bodies. 



153 



Hence, from 

ml'a 2 = [r 2 dm = J (x 2 + if) dm + \R 2 dm — 2h$xdm — 2k]ydm 
we find 

mk 2 — mk g 2 + mR 2 , 
or 

rCa -—fog ~v -tv" > 

It is readily seen that if k a 2 and k^ 2 are the values of k 2 for any 
bod}- with reference to two parallel axes, distant R a and Ri, re- 
spectively from the center of gravity, 




Vt*.o) 






Fig. 117. 



342. Fundamental Radii of Gyration. Straight Line and 
Rectangle. — The squared radius of gyration for a straight line of 
length I with reference to an axis, a, perpendicular to I at one end, 

I 2 

is -=- , and with reference to a parallel to a through the middle 
o 

I 2 
point of I is ^r , for, from the theorem of parallel axes, we have 

72 p 

-— = kg 2 H — -7- . The same theorem will give k 2 for the line I 
o 4 

with reference to a perpendicular to I anywhere in space. 

24 



354: 



The Calculus. 



These are also the values of k 2 with reference to the same axes 
for any rectangle having I as one side, and parallels' to the axis 
of reference as the perpendicular sides. 

The squared radius of gyration for a rectangle having the 
dimensions a and b, with reference to an axis perpendicular to 
its plane through its center, is found from the theorem of per- 
pendicular axes to be — ^ — . 

The theorem of parallel axes will now give the value of k 2 for 
any axis perpendicular to the plane of the rectangle. 

These will also be the values of k 2 , with reference to the same 
axes, for any right parallelopiped having the rectangle as a 
cross-section. 



343. Examples. 

Find 7 and k 2 for each of the following homogeneous figures 
with reference to three mutually perpendicular axes through the 
center of gravity, one of them being an axis of symmetry. 

1. 





2" 


_£ 7,2_11 jL 2 

O, H,y — --g-, ti z — 

= 180, I y 2 = 132, 7 


/=312 (Fig. 118). 

lh 25 H 

2\ 






6 






* 




I 

JO 

1 






3" 




*" 






3 








3 




\ 

5 

i 




F 


8" 

[G. 1 


L18. 




u. v 

F 


72 

LG. 


H 

119. 





8025, 7^=16,280, 7* = 24,305, 

*--*-, k z 2 = ^H¥ig.ll9). 



_5 3_ 

ti 



■L. 2_ 407 

, tly g , 









Eigid Bodies. 


h= 


hb 3 
48 " 


> ■*!/ 


fc 3 6 

~ 36 ' 






4 = 


3^(3& 2 + 4/* 2 ), 


fa f = 


b 2 
"24' 


Ky 


/i 2 
~18' 



355 



k z 2 =^(3b 2 + 4:h 2 ) (Fig. 120). 



Fig. 120. 



4. Find fc 2 for these figures with reference to axes perpendicu- 
lar to their planes, the axis passing through one of the lower 
corners in the figures of examples 1 and 2, through the left vertex 
in the figure of example 3. 

Ans. (1) V=*P, (2) k*=^, (3) P= m 2 , + 6 ' . 

5. One diagonal of a rhombus is equal to a side: find Tc 2 with 

d 2 GL 2 

reference to each diagonal. Ans. ^7 an d ^ ■ 

6. The dimensions of a rectangular parallelopiped are I, a and 
b. Find its moment of inertia with reference to an edge of 

length Z. Ans. ^~(a 2 + b 2 ). 

7. Find the moment of inertia with reference to a lateral edge 
for a right prism of which the cross-section is an equilateral 
triangle. 

Ans. Lateral edge being I, edge of base a, I=-f^la*'\/S. 

8. A cube balanced on one edge on a horizontal plane is slightly 
disturbed and falls without sliding; what is its angular rate of 
rotation when it hits the plane, and what is then the speed of a 
point on the edge opposite to the stationary edge? 

Ans. co 2 =-^V2; v 2 = 3glV~2. 

344. k 2 for a Circumference or for a Circle. — For a circum- 
ference of radius a, with reference to an axis perpendicular to its 
plane at its center, Jc = a; and with reference to any diameter, Tc 
must be the same as with reference to any other diameter. The 



356 



The Calculus. 



value of h 2 for a circumference with reference to any diameter is 
therefore determined by the theorem of perpendicular axes, 

a 2 = k* + Jc 2 , to he lc 2 = ~. 

To find k 2 for a circle of radius a with reference to a perpen- 
dicular to its plane at its center 0, divide the circle into elemen- 
tary rings by concentric circumferences dr 
apart; then, assuming the density =1, 

dm = 2>7rrdr, 




r 

_Jo 



27rr z dr 



2irrdr 



i-va* 



TTOf 



Fig. 121. 



Using the theorem of perpendicular axes, 

as we did for the circumference, we find h 2 for the circle with 

a 2 
reference to any diameter to be — r- . 

We might have obtained the last result by observing that Tc 2 

r 2 . .* 

for the element %irrdr is -=- , so that its momenr of inertia is 

irr z dr, and for the circle, 



F=- J 



xr s dr 



2*rdr 



na 2 



Again, we might have used rectangular coordinates, taking the 
diameter in question as the axis of x, and using either the element 

xdy, for which Jc = y J or the element ydx, for which h 2 = ~~ . The 

corresponding elements of the moment of inertia would have been 
xy 2 dy and iy B dx. 

The theorem of parallel axes will now give h 2 for any circum- 
ference or circle with reference to any axis in or perpendicular to 
its plane. 



Eigid Bodies. 357 

The value of k 2 for a right circular cylinder with reference to 
its geometric axis or to any parallel axis is the same as for its 
right section. 

345. k 2 for any Area or Arc. — The methods suggested for the 
circle will give h 2 for any area with reference to the coordinate 
axes, and, thence, by the theorem of perpendicular axes, with 
reference to a perpendicular to the plane at the origin. The 
theorem of parallel axes will then give k 2 with reference to any 
axis parallel to either coordinate axis or perpendicular to both of 
them. These results will include the values of h 2 for any right 
cylinder with reference to any parallel to its elements. 

If the boundary of an area is given by a polar equation, it is 
best to take dm = rdOdr and perform a double integration, and 
if k 2 is wanted for an axis perpendicular to the plane at the pole, 
to determine it directly. 

346. Examples. 

1. Find k 2 for an ellipse having the semi-axes a and b with 
reference to each principal diameter and to the perpendicular to 
the plane of the ellipse at the center. 

Ans. k x 2 =-^-, ky 2 =-°~- , k z 2 =z°^— . 

2. Find / for an elliptic right cylinder with reference to an 
element through an extremity of the major axis of a right section. 
Length I, semi-axes a (major) and b. 

Ans. lTrabl(5a 2 + V). 

3. Find k 2 for a parabolic segment, height K, base 2b, with 
reference to its axis of symmetry and to a perpendicular to its 
plane through its vertex. 

Ans. k x 2 = ib 2 , k z 2 = -jt{7b 2 + 15a 2 ). 

4. Find k 2 for the area bounded by the #-axis, and an arch of 
the cycloid, x — a{<^> — sin</>), y = a(l — cos<j>), with reference to 
the .r-axis. Ans. k x 2 =^a 2 . 

5. Find k 2 for the area bounded by the cardioid r = 2a sin 2 f. 
with reference to the initial line and to the perpendicular to its 
plane at the pole. Ans. k x 2 —^a 2 , k z 2 =.%^a 2 . 



358 



The Calculus. 



6. Find k 2 for the arc of a cycloidal arch with reference to its 
base. Ans. ffa 2 . 

7. Find k 2 for the arc of a cardioid with reference to a perpen- 
dicular to its plane at the cusp, and with reference to a parallel 
axis through the center of gravity. 

Ans. F=ffa 2 , k g 2 =i^£a\ 

8. Show that if a vertical plane area is submerged in a fluid 
so that its center of gravity is h below the level free surface, its 
center of fluid pressure will be at a depth D below its center of 
gravity, where hD = lcg 2 , k g being the radius of gyration for the 
submerged area with reference to a horizontal axis through its 
center of gravity. 

347. k z for a Solid of Revolution with Reference to the Geo- 
metric Axis. — In finding k 2 for a solid of revolution with refer- 




Fig. 122. 



ence to the axis of revolution, the most convenient element of 
volume is that generated by the revolution of some element of the 
generating area. For example, to find k 2 for a sphere of radius a 
with reference to a diameter, take the diameter as the axis of x; 
the sphere is generated by the revolution about the axis of x of 
the circle x 2 -\-y 2 — (i 2 , or, in polar coordinates, r = a. 
The element ydx generates a disc, for which 



dm = dV = Try 2 dx, 



IC ~ 2 



dl — ^y^dx; 



Eigid Bodies. 359 



(a Ca 

^Trii^dx y 71 \ (a 2 — x 2 ) 2 dx -, 5/1 9 , i\ 02 
2 * = 2 Jo v y = fr*z (!-$ + &) = ?a 

The element xdy generates a cylindrical shell, for which 

dm — dY — 2-rtyxdy, k 2 — y 2 , dl — 2iry z xdy; 

and again, the element rdOdr generates a ring, for which 

dm = dV = 2irr sin OrdOdr, I 2 = r 2 sin 2 6, dl = 2^ sin 3 Odddr; 

each of these, the first by a single integration, the second by a 
double integration, gives in the same way k 2 — %a 2 . 

k 2 for an axis parallel to the axis of revolution is found directly 
by the theorem of parallel axes. 

348. Examples. 

1. Find k 2 for an ellipsoid of revolution with reference to the 
axis of revolution. 

Ans. Length being 2a, greatest transverse diameter 2b, 

k - 5 • 

2. Find k 2 for a cone of revolution with reference to the geo- 
metric axis. 

Ans. Height being h, radius of base b, k 2 =^b 2 . 

3. Find k 2 for a paraboloid of revolution with reference to the 

geometric axis. 

b 2 
Ans. Height being h, radius of base b,k 2 = -=-. 

4. Find k 2 with reference to the geometric axis for the solid 
generated by the cycloid, x=a(<f> — sin <£), y=a(l — cos<£), in 
revolving about its base. Ans. k 2 — 1.575a- 2 . 

5. Find k 2 with reference to the geometric axis for the solid 
generated by revolving the cardioid r = 2asin 2 f about the in- 
itial line. Ans. k 2 —^a 2 . 

6. If all of the solids in examples 1-4 are of the same length, 
and if each of them weighs 480 pounds to the cubic foot and has 
a maximum transverse diameter of 4 feet, find their respective 



360 



The Calculus. 



kinetic energies in foot-pounds when each is making 2 revolutions 

o QPPOT1 H 

Ans. I 1 = 1024«-S I 2 = 384tt 4 , I 3 =:6407r 4 , I 4 = 945tt 4 . 

7. Find & 2 with reference to the geometric axis for the surface 
formed by revolving an arch of the cycloid about its base. 

Ans. & 2 =ffa 2 . 

8. Find h 2 with reference to the geometric axis for the surface 
formed by revolving the cardioid r=2a sm 2 J about the initial 
line. Ans. lc 2 =-\^a 2 . 

349. Axis of Reference not an Axis of Symmetry. — In other 
cases of finding Tc 2 for a solid, it is generally most convenient to 




Fig. 123. 

divide the solid into elements of volume by planes that are either 
parallel or perpendicular to the axis of reference. For example, 
to find h 2 for a cone with reference to a perpendicular to the 
geometric axis through the vertex, divide the cone by planes 
parallel to the axis of reference and perpendicular to the geome- 
tric axis (see Fig. 123) . Let x be the distance of any one of these 
planes from the axis of reference ; then the corresponding element 

of volume is dV = ir(^p) dx, and its h 2 isz 2 + i (-£) . 

The moment of inertia is therefore, if we take p = l, 



Eigid Bodies. 361 



V= ^ h ; hence ]c 2 = Y \(4:h 2 + b 2 ). 
With reference to a parallel axis through the center of gravity, 

350. Examples. 

1. Find A; 2 for a cylinder of altitude h, radius of base ~b, with 

reference to a diameter of its base, and to a parallel axis through 

b 2 h 2 b 2 h 2 

the center of gravity. Ans. k 2 = -r- -f -^- , &/ = -j + y~-. 

2. Find & 2 for a paraboloid of revolution of altitude h, radius 

of base b, with reference to a tangent at the vertex, and to a 

parallel axis through the center of gravity. 

b 2 h 2 b 2 h 2 

Ans. J*=_ + __,V=- g - + ls . 

3. Find k 2 for an ellipsoid of semi-axes a, b, c with reference 
to each of its principal diameters. 

Axu, ,&=»+*, V=^, W=^- 



4. A circle of radius a is revolved about a line in its plane dis- 
tant na(n>l) from its center. Find h 2 with reference to the 
geometric axis and with reference to a diameter perpendicular to 
the geometric axis for the solid so formed, and for its surface. 

a 2 
Ans. Geom. axis: Solid, k 2 — —r- (4n 2 + 3) ; surface, h 2 = 

a 2 a 2 

- 7V -(2n 2 + 3). Perp. axis: Solid, fc 2 = -— - (4n 2 + 5) ; surface, 

k 2 =~(2n 2 + 5). 

5. A parabolic segment, height fe/ base 2b, revolves about the 
base; find h 2 for the solid so generated, with reference to a 
diameter of the maximum circular section and with reference to 
the geometric axis. Ans. k x 2 =^ T (4zh 2 + 3b 2 ), k z 2 =£ T h 2 . 

351. D'Alembert's Principle. — For convenience of statement 
we divide the forces acting on a body or system of bodies into two 
classes : external or impressed, forces, the source of which is out- 



362 The Calculus. 

side of the body, and internal forces, the reactions between the 
particles of the body or system. 

Consider a system composed of two connected particles, P t 
and P 2 , and suppose the total external force acting on P t to be f lf 
and that acting on P 2 to be f 2 . Let the reaction of P. 2 upon P x 
be r 1} then the reaction of P x upon P 2 is r 2 = — r t , a force having 
the same line of action as r ± . Let the accelerations of P ± and P 2 
be a x and a 2 , their masses m x and m 2 . Then m^ is the single 
force which would give P t the motion it actually has; ra^ is 
therefore called the effective force for P lm m 2 a 2 is the effective 
force for P 2 . Now the equations of motion for P ± and P 2 are : 

/i + r i = w i a i> f 2 + r 2 = m 2 a 2 , 



Fig. 124. 

where the + indicates the process of finding a resultant, and the 
equality of (f + r) with ma holds for magnitude, direction and 
line of action ; i. e., is an identity. If we now add these identities, 
we have 

A + r ± + f 2 + r 2 = m 1 a 1 + ra 2 a 2 , 

which reduces, on account of the relation between r x and r 2 , to 

fl + fi — W l a l + ^2 a 2? 

another identity. 

This reasoning can evidently be applied to any number of par- 
ticles, for all the internal forces occur in equal, directly opposed 
pairs, and drop out in the summation. If the particles form a 
continuous body or system of such bodies, the summation becomes 



Eigid Bodies. 363 

an integration. Then D'Alembert's Principle may be stated as 
follows : 

The resultant of the impressed forces acting on a material 
body or system is identical with the resultant of the effective 
forces for all its particles. 

It follows from this that the sum of the resolved parts of the 
impressed forces in any given direction is equal to the sum of the 
resolved parts of the effective forces in the same direction. 

For instance, if a body weighing W pounds is falling under the 
action of gravity alone, we know that the resultant of the im- 
pressed forces is a vertical force of W pounds acting through its 
center of gravity; hence the resultant of the effective forces for 
all its particles, which we may indicate by \adm taken throughout 
the body, is this same force ; then if the body is rotating, so that 
a is not the same for all the particles, the horizontal components 
of adm must balance one another when summed up for the whole 
body. 

352. D'Alembert's Principle Applied to Rotation. — The appli- 
cation to linear motion is only half the use of D'Alembert's Prin- 
ciple ; the other half is its application to angular motion, or rota- 
tion about an axis. The kinematics and dynamics of angular 
motion can be founded theoretically on the corresponding sciences 
of linear motion, or may be established independently. 

It is found that the moment of a force plays the same part in 
angular motion that the force itself plays in linear motion. In 
Art. 314 we defined the moment of a force about an axis perpen- 
dicular to its line of action. If we have a force F and an axis I 
in any position, we can draw a plane parallel to I through F's 
line of action, and resolve F in this plane into f parallel to I and 
/ perpendicular to I. Then the moment of F about I is denned to 
be the same as the moment of / about I. 

It can be shown that the moment about any axis of the resultant 
of anv set of forces is the sum of the moments about the same axis 



364 The Calculus. 

of the component forces. Consequently, D'Alembert's Principle 
tells ns that the sum of the moments of the impressed forces about 
any axis is equal to the sum of the moments of the effective forces 
about the same axis. 

For instance, if a falling body is acted upon by both gravity 
and atmospheric resistance and descends without rotation, all of 
its particles having the same acceleration a, the sum of the 
moments of the effective forces about an axis through its center 
of gravity G, if x is the distance of any particle of mass dm from 
G, is \xadm—a\xdm taken throughout the body, and is therefore 
zero. Moreover, as the sum of the forces of gravity acting on the 
particles passes through G, gravity has a zero moment about the 
same axis; therefore the total atmospheric resistance has the 
moment zero about any axis through G, and so is a single force 
acting through G-. 

If, on the other hand, the falling body starts at rest and rotates 
as it falls, the total atmospheric resistance does not pass through G. 

353. The Equation of Rotation. — If a body is capable of turn- 
ing about a fixed axis, it will turn if acted upon by a force that 
has a moment about that axis. The numerical relation between 
the moment and the rotary motion is obtained as follows: Let 
Fig. 125 represent a section of the body by 
a plane perpendicular to the axis at 0, and 
let dm> at P be an element of mass. Let <a 
be the angular rate at which P moves about 
when the body rotates, and let OP — r; 
then the acceleration of P has two resolved 

v 2 
parts : = no 2 in the direction. PO, and 

— jr — r -=— in the direction of the tangent 

PT. 

The effective force for the particle at P 
is therefore the resultant of r<a 2 dm along PO, which has no mo- 




Pig. 125. 



Rigid Bodies. 365 

ment about the axis, and r -^- dm, which has the moment 

at 

r — — dm. 

at 



Ihe sum of the moments of all the effective forces about the 

dt 



axis of rotation is therefore jr 2 —=r- dm taken throughout the body. 



If the body is rigid, -j— is the same for all its points, and the 
sum becomes 

dt ^ am - L dt - mlc dt ' 

If M is the sum of the moments about the axis of rotation of 
all the external forces, we have from D'Alembert's Principle, 

at 

This is a second equation of motion; it is distinguished from 
the equation of linear motion, f=ma, by being called the equation 
of rotary motion. The two equations are also called the equations 
of translation and rotation. 

There is an evident correspondence in the two equations be- 
tween momentum and force, moment of inertia and mass, angular 
acceleration and linear acceleration. The moment of inertia, 
however, is not, like the mass, a fixed characteristic of the body, 
for it varies with the relative position of the body and the axis. 

354. The Compound Pendulum. — The compound pendulum in 
its simplest form consists of a rigid body capable of turning freely 
on a fixed horizontal axis; this is the pendulum used in clocks and 
in physical apparatus for determining local values of the accelera- 
tion due to gravity. The devices used to minimize friction at the 
axis are so successful that the action of this force may be neg- 



366 



The Calculus. 






lected in considering the motion. The only impressed forces, 
then, are the forces of gravity on the particles of the pendulum, 
the resultant of which acts vertically downward through the 
center of gravity G of the pendulum. 

Let Fig. 126 represent a section of the 
pendulum by a plane perpendicular to 
the axis at and passing through G; OV 
and OG are the traces of planes contain- 
ing the axis, the first of them vertical. 
Let OG = a; then the moment of the im- 
pressed forces is Wa sin 0, if W is the 
weight of the pendulum and the angle 
YOG is called 6. 
According to Art. 353, 

doj 




Fig. 126. 



Wa sin 0: 



la 



dt 



W k* ** 

9 a dt 2 



and I a and k a 2 are the moment of inertia and squared radius of 
gyration for the pendulum with reference to the axis of rotation. 
Hence 



d 2 6 _ 
dt 2 ~ 



k 2 



sin 6. 



This is precisely the equation for the motion of a simple pen- 

k 2 
dulum of which the length is 1= —?— ; that is, the compound 

pendulum vibrates as if all its mass were concentrated at a point 
Q of the line OG, called the center of oscillation, at a distance I 
from 0. lis called the length of the equivalent simple pendulum. 
If the squared radius of gyration for the pendulum with refer- 
ence to a parallel to the axis of rotation through G is k g 2 , 



fCg 



a 



If we let GQ = b, OQ = a+b and 



ab 



Eigid Bodies. 367 

355. Examples. 

1. A cone of altitude 4a, base of radius 2a, is free to rotate 
about a small smooth axis, perpendicular to the geometric axis, a 
from the vertex. Find the length of the equivalent simple 
pendulum. Ans. l—^-a. 

2. A rod of a centrifugal governor is \ inch in diameter and 
6 inches long and carries a sphere 1 inch in diameter at the lower 
end. What is the length of the equivalent simple pendulum, 
supposing the rod suspended from its extreme upper end ? 

Ans. Very nearly 6 inches. 

356. Moments of Inertia Determined by Experiment. — Sup- 
pose we wish to find the moment of inertia of a body with refer- 
ence to a given axis, and that the shape is inconvenient for com- 
putation. An apparatus may be set up similar to a lathe, but 
driven by a weight hung from a light cord wound on a drum 
carried on the axle and concentric with its axis of revolution. 
The body may be fastened to both head-stocks in such a way that 
the axis with reference to which the moment of inertia is desired 
will coincide with the axis of revolution. Let the radius of the 
drum be \ foot and the driving weight 10 pounds, and suppose the 
weight is observed to fall \\ feet in the first 2 seconds. Suppose 
the moment of inertia of the moving parts of the apparatus itself 
to be -2V (engineer's units) and assume the effect of friction 
negligible. Let g — Z2. 

Let the weight descend 5 feet, and the apparatus rotate through 
6 radians, in t seconds, and let T be the tension in the cord. Then 
the equation of translation for the weight and the equation of 
rotation for the revolving mass give, if I is the moment of inertia 
of the body and the apparatus together, 

10 ~ T =™W> (1) 

Txi=I§. (2) 



368 The Calculus. 



l^=2^f,so,from(2), 



dt 2 dt 

■A _ 

dt 



d 2 <i 



and from (1), 



10-4a7=Ha, I=rA(32-a). 



The constant acceleration a is readily seen to be f f /s 2 ; hence 
/ = f |f. The moment of inertia of the body alone is T =fH — 2V 
If the body weighs 150 pounds, the corresponding radius of gyra- 
tion is k, where 

F = §)' =M-3f£<r = ff(l-0.01536), 
& = f (1-0.008) feet = 4.96 inches. 

If the effect of friction cannot be neglected, it is best allowed 
for by making two tests with different driving weights and elim- 
inating the moment of friction and the two tensions from the 
four equations of motion. 

If a body is suspended from an axis and allowed to vibrate 

through a small angle, its time of vibration, T — ir J -, will give 

aT 2 k 2 
the length of the equivalent simple pendulum, I — ■^- r - = , 

where k is the radius of gyration with reference to the axis of 
suspension, and a is the distance of this axis from the center of 

gravity. Then if a is known, k— — V ga. 

357. Combined Translation and Rotation. — It can be shown 
that when a body moves in any way, its center of gravity moves 
as if it were a heavy particle having the same mass as the body 
and acted upon by forces equal to the forces acting on the body. 
At the same time, the body if rigid revolves about any axis 



Kigid Bodies. 



369 



through its center of gravity precisely as it would if the center of 
gravity were fixed. 

Consider, for instance, the motion of 
a sphere down a rough inclined plane. 
Let the weight of the sphere be W 
pounds, its radius a feet, the coefficients 
of friction /x and / and the inclination 
of the plane to the horizontal <j>. In t 
seconds, let the center of the sphere 
move s feet down the plane, and let the 
radius to the point initially in contact 
with the plane rotate through radians. 

Then if the sphere rolls, s — aB; if it also slides, s>a$. The roll- 
ing is due to rotation about the horizontal axis, and is caused by 
friction alone ; the friction may not be sufficient to cause rotation 
rapid enough to keep up with the translation. 

The equation for translation down the plane gives 




Fig. 127. 



w ■ jl j? W d 2 s 

W sm<£ — F— — -jjz 

g dt 2 



d~s 



The equation for rotation about the horizontal diameter gives 



dt 



aF = I^ =fa 2 



w <m 
g dt 2 



d 2 l 
dt 2 



5g F 
2a W 



d 2 s d 2 

If the sphere merely rolls, s — aO,-^ =a -^ , 

Qi 1 



/ • F 

g(sm<f>- w 



F -§sm<f>. 



Thence 



d 2 s * • , 

-3P =?♦«"*• 



(3) 



(4) 



From this constant acceleration, the motion of the center of the 
sphere down the plane can be determined directly. 



370 



The Calculus. 



To determine whether pure rolling is possible or not, we have 
the equation for translation normal to the plane : 

R=W cos <f>, . (5) 

■pi 
from which ~» =y tan*, by (3). The greatest value possible 

for-p- i s ht the coefficient of statical friction; if fi is greater than 

f tan *, the sphere will merely roll ; if fi is less than f tan *, the 
sphere will also slide. 

In case the sphere both rolls and slides, we have the equations 

■pp 
of motion, 1, 2 and 5, and we also know that -^- = // constantly. 

Thus we have 



^=2^ cob* 



d 2 
it 2 ' 
dO 

dt 



(fVcos*)*. 

, = (|L,, C08+ ),. 



v = 



-^=#(sin*-//cos*). 
dt 



(sin* — /*' cos <f>)t. 
s = ^g(sm<f> — fx cos*)£ 2 



The amount of slipping at any time is (s — a6), and the rate of 
slipping is v — a*) = gt (am <j>—i(i cos *) f/s. 



358. 



Examples. 



1. Show that if a cylinder rolls directly down an inclined plane 
without sliding, //, must be greater than -J tan *, and the accelera- 
tion down the plane is § g sin * f/s 2 . Show that if the cylinder 
slides as well as rolls, the acceleration of the center down the 
plane is g (sin * — /*' cos *)f/s 2 , and the rate at which the point of 
contact slips on the surface is gt (am * — 3// cos *) f/s. 

2. Two spheres, each of radius a and weighing W pounds, 
look alike, and will stand in any position on a horizontal plane, 
but one is said to be solid, the other hollow. They are rolled 
(without sliding) down an inclined plane; sphere A goes 10 feet 



Eigid Bodies. 371 

and sphere B 9 feet in the same time. Which is hollow, and what 
is its radius of gyration ? 

Ans. -172- = 2 4- jl2 m g enera l> B is hollow, with l ,2 -4 a 2 - 

3. A hollow cylinder of mean radius a, of which the thickness 
may be neglected, and a homogeneous cylinder of radius a having 
the same weight are started together down an inclined plane; 
show that their accelerations are in the ratio j if they roll without 
sliding. 



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